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Section 1.4
                 Calculating Limits

                    V63.0121.041, Calculus I

                         New York University


                      September 15, 2010


Announcements

   First written homework due today (put it in the envelope)
   Remember to put your lecture and recitation section numbers on
   your paper

                                               .   .   .   .   .   .
Announcements




         First written homework due
         today (put it in the
         envelope) Remember to
         put your lecture and
         recitation section numbers
         on your paper




                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       2 / 45
Yoda on teaching a concepts course

“You must unlearn what you have learned.”




In other words, we are building up concepts and allowing ourselves
only to speak in terms of what we personally have produced.
                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       3 / 45
Objectives


         Know basic limits like
         lim x = a and lim c = c.
         x→a                      x→a
         Use the limit laws to
         compute elementary limits.
         Use algebra to simplify
         limits.
         Understand and state the
         Squeeze Theorem.
         Use the Squeeze Theorem
         to demonstrate a limit.



                                                                         .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010       4 / 45
Outline


Recall: The concept of limit

Basic Limits

Limit Laws
   The direct substitution property

Limits with Algebra
   Two more limit theorems

Two important trigonometric limits


                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       5 / 45
Heuristic Definition of a Limit



Definition
We write
                                       lim f(x) = L
                                      x→a

and say

                   “the limit of f(x), as x approaches a, equals L”

if we can make the values of f(x) arbitrarily close to L (as close to L as
we like) by taking x to be sufficiently close to a (on either side of a) but
not equal to a.




                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       6 / 45
The error-tolerance game


A game between two players (Dana and Emerson) to decide if a limit
lim f(x) exists.
x→a
Step 1 Dana proposes L to be the limit.
Step 2 Emerson challenges with an “error” level around L.
Step 3 Dana chooses a “tolerance” level around a so that points x
       within that tolerance of a (not counting a itself) are taken to
       values y within the error level of L. If Dana cannot, Emerson
       wins and the limit cannot be L.
Step 4 If Dana’s move is a good one, Emerson can challenge again or
       give up. If Emerson gives up, Dana wins and the limit is L.



                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       7 / 45
The error-tolerance game




  L
  .




        .
                                     a
                                     .

      To be legit, the part of the graph inside the blue (vertical) strip
      must also be inside the green (horizontal) strip.
      If Emerson shrinks the error, Dana can still win.
                                                                   .   .    .      .      .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       8 / 45
Limit FAIL: Jump
                                              y
                                              .



                                           . .
                                           1


                                               .                                          x
                                                                                          .

                                                                   .
                                                                   Part of graph
                                        . 1.
                                        −                          inside blue is not
                                                                   inside green




                                                                     .    .    .      .       .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits              September 15, 2010       9 / 45
Limit FAIL: Jump
                                              y
                                              .

          .
          Part of graph
          inside blue is not               . .
                                           1
          inside green

                                               .                                       x
                                                                                       .


                                        . 1.
                                        −




                                                                   .   .    .      .       .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       9 / 45
Limit FAIL: Jump
                                              y
                                              .

          .
          Part of graph
          inside blue is not               . .
                                           1
          inside green

                                               .                                       x
                                                                                       .


                                        . 1.
                                        −



               |x|
      So lim       does not exist.
            x→0 x
                                                                   .   .    .      .       .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010       9 / 45
Limit FAIL: unboundedness
                           y
                           .

                                     .
                                          1
                                         lim+
                                            does not exist
                                     x→0 x
                                     because the function
                                     is unbounded near 0

                       .?.
                       L




                            .                                                              x
                                                                                           .
                                0
                                .


                                                                     .   .     .      .        .   .

 V63.0121.041, Calculus I (NYU)     Section 1.4 Calculating Limits           September 15, 2010    10 / 45
Limit EPIC FAIL
                                                          (π )
Here is a graph of the function f(x) = sin                         :
                                                             x
                                            y
                                            .
                                          . .
                                          1


                                              .                                       x
                                                                                      .


                                       . 1.
                                       −

For every y in [−1, 1], there are infinitely many points x arbitrarily close
to zero where f(x) = y. So lim f(x) cannot exist.
                                  x→0


                                                                       .   .    .         .   .     .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits               September 15, 2010   11 / 45
Outline


Recall: The concept of limit

Basic Limits

Limit Laws
   The direct substitution property

Limits with Algebra
   Two more limit theorems

Two important trigonometric limits


                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   12 / 45
Really basic limits



Fact
Let c be a constant and a a real number.
 (i) lim x = a
       x→a
 (ii) lim c = c
       x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   13 / 45
Really basic limits



Fact
Let c be a constant and a a real number.
 (i) lim x = a
       x→a
 (ii) lim c = c
       x→a


Proof.
The first is tautological, the second is trivial.




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   13 / 45
ET game for f(x) = x

                  y
                  .




                   .           x
                               .




                       .   .       .   .   .   .
ET game for f(x) = x

                  y
                  .




                   .           x
                               .




                       .   .       .   .   .   .
ET game for f(x) = x

                  y
                  .



                 . .
                 a




                   .    .           x
                                    .
                       a
                       .




                            .   .       .   .   .   .
ET game for f(x) = x

                  y
                  .



                 . .
                 a




                   .    .           x
                                    .
                       a
                       .




                            .   .       .   .   .   .
ET game for f(x) = x

                  y
                  .



                 . .
                 a




                   .    .           x
                                    .
                       a
                       .




                            .   .       .   .   .   .
ET game for f(x) = x

                                   y
                                   .



                                  . .
                                  a




                                    .                              .             x
                                                                                 .
                                                                  a
                                                                  .




                                                                         .   .         .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits               September 15, 2010   14 / 45
ET game for f(x) = x

                                   y
                                   .



                                  . .
                                  a




                                    .                              .             x
                                                                                 .
                                                                  a
                                                                  .



      Setting error equal to tolerance works!
                                                                         .   .         .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits               September 15, 2010   14 / 45
ET game for f(x) = c




                   .




                       .   .   .   .   .   .
ET game for f(x) = c

                  y
                  .




                   .           x
                               .




                       .   .       .   .   .   .
ET game for f(x) = c

                  y
                  .




                   .           x
                               .




                       .   .       .   .   .   .
ET game for f(x) = c

                     y
                     .



                     .
                 c
                 .




                     .    .           x
                                      .
                         a
                         .




                              .   .       .   .   .   .
ET game for f(x) = c

                     y
                     .



                     .
                 c
                 .




                     .    .           x
                                      .
                         a
                         .




                              .   .       .   .   .   .
ET game for f(x) = c

                     y
                     .



                     .
                 c
                 .




                     .    .           x
                                      .
                         a
                         .




                              .   .       .   .   .   .
ET game for f(x) = c

                                      y
                                      .



                                      .
                                  c
                                  .




                                      .                              .             x
                                                                                   .
                                                                    a
                                                                    .



      any tolerance works!
                                                                           .   .         .      .      .    .

 V63.0121.041, Calculus I (NYU)           Section 1.4 Calculating Limits               September 15, 2010   15 / 45
Really basic limits



Fact
Let c be a constant and a a real number.
 (i) lim x = a
       x→a
 (ii) lim c = c
       x→a


Proof.
The first is tautological, the second is trivial.




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   16 / 45
Outline


Recall: The concept of limit

Basic Limits

Limit Laws
   The direct substitution property

Limits with Algebra
   Two more limit theorems

Two important trigonometric limits


                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   17 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   18 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   18 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   19 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M
      x→a
 3. lim [cf(x)] = cL
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   19 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M
      x→a
 3. lim [cf(x)] = cL (error scales)
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   19 / 45
Justification of the scaling law



      errors scale: If f(x) is e away from L, then

                              (c · f(x) − c · L) = c · (f(x) − L) = c · e

      That is, (c · f)(x) is c · e away from cL,
      So if Emerson gives us an error of 1 (for instance), Dana can use
      the fact that lim f(x) = L to find a tolerance for f and g
                           x→a
      corresponding to the error 1/c.
      Dana wins the round.




                                                                        .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)        Section 1.4 Calculating Limits           September 15, 2010   20 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling)
      x→a
 3. lim [cf(x)] = cL (error scales)
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   21 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling)
      x→a
 3. lim [cf(x)] = cL (error scales)
      x→a
 4. lim [f(x)g(x)] = L · M
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   22 / 45
Limits and arithmetic



Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
              x→a                 x→a
 1. lim [f(x) + g(x)] = L + M (errors add)
      x→a
 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling)
      x→a
 3. lim [cf(x)] = cL (error scales)
      x→a
 4. lim [f(x)g(x)] = L · M (more complicated, but doable)
      x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   22 / 45
Limits and arithmetic II


Fact (Continued)
             f(x)  L
 5. lim           = , if M ̸= 0.
      x→a    g(x)  M




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   23 / 45
Caution!


      The quotient rule for limits says that if lim g(x) ̸= 0, then
                                                              x→a

                                        f(x)    limx→a f(x)
                                  lim        =
                                  x→a   g(x)   limx→a g(x)

      It does NOT say that if lim g(x) = 0, then
                                   x→a

                                      f(x)
                                  lim      does not exist
                                  x→a g(x)

      In fact, limits of quotients where numerator and denominator both
      tend to 0 are exactly where the magic happens.
      more about this later

                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits            September 15, 2010   24 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [        ]n
              n
 6. lim [f(x)] = lim f(x)
      x→a                    x→a




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [       ]n
              n
 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
      x→a                    x→a




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [       ]n
              n
 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
      x→a                    x→a
              n        n
 7. lim x = a
      x→a




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [       ]n
              n
 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
      x→a                    x→a
              n        n
 7. lim x = a
    x→a
        √     √
 8. lim n x = n a
      x→a




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [       ]n
              n
 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
      x→a                    x→a
              n        n
 7. lim x = a (follows from 6)
    x→a
        √     √
 8. lim n x = n a
      x→a




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Limits and arithmetic II


Fact (Continued)
         f(x)   L
 5. lim       = , if M ̸= 0.
        g(x)
      x→a       M
                 [       ]n
              n
 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
      x→a                    x→a
              n        n
 7. lim x = a (follows from 6)
    x→a
        √     √
 8. lim n x = n a
    x→a
        √        √
 9. lim n f(x) = n lim f(x) (If n is even, we must additionally assume
      x→a                         x→a
      that lim f(x) > 0)
             x→a




                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   25 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   26 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3


Solution
By applying the limit laws repeatedly:
              (             )
          lim x2 + 2x + 4
               x→3




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   26 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3


Solution
By applying the limit laws repeatedly:
              (             )       ( )
          lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
               x→3                       x→3                  x→3           x→3




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits            September 15, 2010   26 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3


Solution
By applying the limit laws repeatedly:
              (             )       ( )
          lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
          x→3                   x→3       x→3       x→3
                                (      )2
                              = lim x + 2 · lim (x) + 4
                                            x→3                    x→3




                                                                   .     .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   26 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3


Solution
By applying the limit laws repeatedly:
              (             )       ( )
          lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
          x→3                   x→3       x→3       x→3
                                (      )2
                              = lim x + 2 · lim (x) + 4
                                            x→3                    x→3

                                     = (3)2 + 2 · 3 + 4



                                                                   .     .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   26 / 45
Applying the limit laws

Example
        (           )
Find lim x2 + 2x + 4 .
       x→3


Solution
By applying the limit laws repeatedly:
              (             )       ( )
          lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
          x→3                   x→3       x→3       x→3
                                (      )2
                              = lim x + 2 · lim (x) + 4
                                            x→3                    x→3

                                     = (3)2 + 2 · 3 + 4
                                     = 9 + 6 + 4 = 19.

                                                                   .     .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   26 / 45
Your turn




Example
           x2 + 2x + 4
Find lim
       x→3   x3 + 11




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   27 / 45
Your turn




Example
           x2 + 2x + 4
Find lim
       x→3   x3 + 11

Solution
                       19  1
The answer is             = .
                       38  2




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   27 / 45
Direct Substitution Property




Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of f, then

                                    lim f(x) = f(a)
                                    x→a




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   28 / 45
Outline


Recall: The concept of limit

Basic Limits

Limit Laws
   The direct substitution property

Limits with Algebra
   Two more limit theorems

Two important trigonometric limits


                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   29 / 45
Limits do not see the point! (in a good way)

Theorem
If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L.
                                       x→a                             x→a




                                                                   .    .      .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   30 / 45
Limits do not see the point! (in a good way)

Theorem
If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L.
                                       x→a                             x→a


Example
            x2 + 2x + 1
Find lim                , if it exists.
       x→−1    x+1




                                                                   .    .      .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   30 / 45
Limits do not see the point! (in a good way)

Theorem
If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L.
                                       x→a                             x→a


Example
            x2 + 2x + 1
Find lim                , if it exists.
       x→−1    x+1

Solution
      x2 + 2x + 1
Since             = x + 1 whenever x ̸= −1, and since
         x+1
                             x2 + 2x + 1
 lim x + 1 = 0, we have lim              = 0.
x→−1                    x→−1    x+1

                                                                   .    .      .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits             September 15, 2010   30 / 45
x2 + 2x + 1
ET game for f(x) =
                      x+1

                                y
                                .




                         .       .                x
                                                  .
                       −
                       . 1




   Even if f(−1) were something else, it would not effect the limit.

                                              .    .   .    .    .     .
x2 + 2x + 1
ET game for f(x) =
                      x+1

                                               y
                                               .




                                    .           .                        x
                                                                         .
                                  −
                                  . 1




      Even if f(−1) were something else, it would not effect the limit.

                                                                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)     Section 1.4 Calculating Limits           September 15, 2010   31 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                   .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                         .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                          MTP                   DSP
                                  lim+ f(x) =      lim+ x2 = 02 = 0
                                  x→0             x→0



                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                         .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                          MTP                   DSP
                                  lim+ f(x) =      lim+ x2 = 02 = 0
                                  x→0             x→0



                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                         .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                          MTP                   DSP
                                  lim+ f(x) =      lim+ x2 = 02 = 0
                                  x→0             x→0



                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                          .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                           MTP                   DSP
                                  lim+ f(x) =       lim+ x2 = 02 = 0
                                  x→0              x→0

Likewise:
                                   lim f(x) = lim −x = −0 = 0
                                  x→0−           x→0−                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)          Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                          .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                           MTP                   DSP
                                  lim+ f(x) =       lim+ x2 = 02 = 0
                                  x→0              x→0

Likewise:
                                   lim f(x) = lim −x = −0 = 0
                                  x→0−           x→0−                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)          Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Limit of a function defined piecewise at a boundary
point

Example
Let
                {
                 x2 x ≥ 0                                                          .
         f(x) =
                 −x x < 0

Does lim f(x) exist?
         x→0

Solution
We have
                                           MTP                   DSP
                                  lim+ f(x) =       lim+ x2 = 02 = 0
                                  x→0              x→0

Likewise:
                                   lim f(x) = lim −x = −0 = 0
                                  x→0−           x→0−                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)          Section 1.4 Calculating Limits           September 15, 2010   32 / 45
Finding limits by algebraic manipulations


Example
         √
          x−2
Find lim       .
     x→4 x − 4




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   33 / 45
Finding limits by algebraic manipulations


Example
         √
          x−2
Find lim       .
     x→4 x − 4


Solution
                                              √ 2       √       √
Write the denominator as x − 4 =               x − 4 = ( x − 2)( x + 2).




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   33 / 45
Finding limits by algebraic manipulations


Example
         √
          x−2
Find lim       .
     x→4 x − 4


Solution
                                 √ 2        √       √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So
                    √                   √
                      x−2                x−2
                lim         = lim √         √
                x→4 x − 4      x→4 ( x − 2)( x + 2)
                                      1     1
                            = lim √       =
                               x→4   x+2    4



                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   33 / 45
Your turn

Example
Let
                 {
                     1 − x2       x≥1
       f(x) =
                     2x           x<1

Find lim f(x) if it exists.
       x→1




                                                                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)     Section 1.4 Calculating Limits           September 15, 2010   34 / 45
Your turn

Example
Let
                 {
                     1 − x2        x≥1
       f(x) =
                     2x            x<1

Find lim f(x) if it exists.
       x→1

Solution
We have
                                                  (      )
                                                           DSP
                                  lim+ f(x) = lim+ 1 − x2 = 0
                                  x→1           x→1




                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   34 / 45
Your turn

Example
Let
                 {
                     1 − x2        x≥1
       f(x) =
                     2x            x<1
                                                                             .           .
Find lim f(x) if it exists.                                                            1
                                                                                       .
       x→1

Solution
We have
                                                  (      )
                                                           DSP
                                  lim+ f(x) = lim+ 1 − x2 = 0
                                  x→1           x→1




                                                                         .   .     .         .   .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   34 / 45
Your turn

Example
Let
                 {
                     1 − x2        x≥1
       f(x) =
                     2x            x<1
                                                                             .           .
Find lim f(x) if it exists.                                                            1
                                                                                       .
       x→1

Solution
We have
                                                  (      )
                                                           DSP
                                  lim+ f(x) = lim+ 1 − x2 = 0
                                  x→1           x→1
                                                                  DSP
                                  lim f(x) = lim (2x) = 2
                                    −                 −
                                  x→1           x→1

                                                                         .   .     .         .   .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   34 / 45
Your turn

Example
                                                                                        .
Let
                 {
                     1 − x2        x≥1
       f(x) =
                     2x            x<1
                                                                             .           .
Find lim f(x) if it exists.                                                            1
                                                                                       .
       x→1

Solution
We have
                                                  (      )
                                                           DSP
                                  lim+ f(x) = lim+ 1 − x2 = 0
                                  x→1           x→1
                                                                  DSP
                                  lim f(x) = lim (2x) = 2
                                    −                 −
                                  x→1           x→1

                                                                         .   .     .         .   .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   34 / 45
Your turn

Example
                                                                                        .
Let
                 {
                     1 − x2        x≥1
       f(x) =
                     2x            x<1
                                                                             .           .
Find lim f(x) if it exists.                                                            1
                                                                                       .
       x→1

Solution
We have
                                                  (      )
                                                           DSP
                                  lim+ f(x) = lim+ 1 − x2 = 0
                                  x→1           x→1
                                                                  DSP
                                  lim f(x) = lim (2x) = 2
                                    −                 −
                                  x→1           x→1

The left- and right-hand limits disagree, so the limit does not exist.   .   .     .         .   .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   34 / 45
A message from the Mathematical Grammar Police




Please do not say “ lim f(x) = DNE.” Does not compute.
                                  x→a




                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   35 / 45
A message from the Mathematical Grammar Police




Please do not say “ lim f(x) = DNE.” Does not compute.
                                  x→a
      Too many verbs




                                                                         .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   35 / 45
A message from the Mathematical Grammar Police




Please do not say “ lim f(x) = DNE.” Does not compute.
                                  x→a
      Too many verbs
      Leads to FALSE limit laws like “If lim f(x) DNE and lim g(x) DNE,
                                                           x→a                         x→a
      then lim (f(x) + g(x)) DNE.”
              x→a




                                                                         .   .     .         .   .    .

 V63.0121.041, Calculus I (NYU)         Section 1.4 Calculating Limits           September 15, 2010   35 / 45
Two More Important Limit Theorems

Theorem
If f(x) ≤ g(x) when x is near a (except possibly at a), then

                                    lim f(x) ≤ lim g(x)
                                    x→a              x→a

(as usual, provided these limits exist).

Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a),
and
                         lim f(x) = lim h(x) = L,
                                  x→a           x→a

then
                                        lim g(x) = L.
                                        x→a

                                                                     .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)     Section 1.4 Calculating Limits           September 15, 2010   36 / 45
Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressions
with simple ones when taking the limit.




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   37 / 45
Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressions
with simple ones when taking the limit.
Example
                                  (π )
Show that lim x2 sin                     = 0.
                x→0                x




                                                                           .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)           Section 1.4 Calculating Limits           September 15, 2010   37 / 45
Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressions
with simple ones when taking the limit.
Example
                                  (π )
Show that lim x2 sin                     = 0.
                x→0                x

Solution
We have for all x,
                                  (π )                                         (π )
                  −1 ≤ sin               ≤ 1 =⇒ −x2 ≤ x2 sin                          ≤ x2
                                    x                                           x
The left and right sides go to zero as x → 0.


                                                                           .    .       .      .      .    .

 V63.0121.041, Calculus I (NYU)           Section 1.4 Calculating Limits              September 15, 2010   37 / 45
Illustration of the Squeeze Theorem



                                  y
                                  .                                        . (x) = x2
                                                                           h




                                  .                                        x
                                                                           .




                                                                       .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010   38 / 45
Illustration of the Squeeze Theorem



                                  y
                                  .                                        . (x) = x2
                                                                           h




                                  .                                        x
                                                                           .




                                                                           .(x) = −x2
                                                                           f




                                                                       .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010   38 / 45
Illustration of the Squeeze Theorem



                                  y
                                  .                                        . (x) = x2
                                                                           h
                                                                                             (π )
                                                                           . (x) = x2 sin
                                                                           g
                                                                                                   x

                                  .                                        x
                                                                           .




                                                                           .(x) = −x2
                                                                           f




                                                                       .   .     .      .      .       .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010      38 / 45
Outline


Recall: The concept of limit

Basic Limits

Limit Laws
   The direct substitution property

Limits with Algebra
   Two more limit theorems

Two important trigonometric limits


                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   39 / 45
Two important trigonometric limits




Theorem
The following two limits hold:
         sin θ
     lim        =1
    θ→0 θ
         cos θ − 1
     lim           =0
    θ→0       θ




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   40 / 45
Proof of the Sine Limit

Proof.
                                                              Notice

                                                                               θ




                                  .
                                  θ
              .
              θ
     .
                                      1
                                      .




                                                                           .       .     .      .      .    .

 V63.0121.041, Calculus I (NYU)           Section 1.4 Calculating Limits               September 15, 2010   41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice

                                                               sin θ ≤ θ




                      . in θ .
                      s      θ
              .
              θ
     .
              c
              . os θ              1
                                  .




                                                                       .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010   41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice

                                                               sin θ ≤ θ                     tan θ




                      . in θ . .an θ
                      s      θ t
              .
              θ
     .
              c
              . os θ              1
                                  .




                                                                       .   .     .      .      .     .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010    41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice
                                                                                     θ
                                                               sin θ ≤ θ ≤ 2 tan       ≤ tan θ
                                                                                     2




                      . in θ . .an θ
                      s      θ t
              .
              θ
     .
              c
              . os θ              1
                                  .




                                                                       .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits           September 15, 2010   41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice
                                                                                      θ
                                                               sin θ ≤ θ ≤ 2 tan        ≤ tan θ
                                                                                      2
                                                          Divide by sin θ:

                                                                              θ       1
                                                                       1≤         ≤
                                                                            sin θ   cos θ
                      . in θ . .an θ
                      s      θ t
              .
              θ
     .
              c
              . os θ              1
                                  .




                                                                       .    .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits            September 15, 2010   41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice
                                                                                      θ
                                                               sin θ ≤ θ ≤ 2 tan        ≤ tan θ
                                                                                      2
                                                          Divide by sin θ:

                                                                              θ       1
                                                                       1≤         ≤
                                                                            sin θ   cos θ
                      . in θ . .an θ
                      s      θ t
              .
              θ                                             Take reciprocals:
     .
              c
              . os θ              1
                                  .                                         sin θ
                                                                       1≥         ≥ cos θ
                                                                              θ



                                                                       .    .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits            September 15, 2010   41 / 45
Proof of the Sine Limit

Proof.
                                                          Notice
                                                                                      θ
                                                               sin θ ≤ θ ≤ 2 tan        ≤ tan θ
                                                                                      2
                                                          Divide by sin θ:

                                                                              θ       1
                                                                       1≤         ≤
                                                                            sin θ   cos θ
                      . in θ . .an θ
                      s      θ t
              .
              θ                                             Take reciprocals:
     .
              c
              . os θ              1
                                  .                                         sin θ
                                                                       1≥         ≥ cos θ
                                                                              θ

As θ → 0, the left and right sides tend to 1. So, then, must the middle
expression.
                                                                       .    .     .      .      .    .

 V63.0121.041, Calculus I (NYU)       Section 1.4 Calculating Limits            September 15, 2010   41 / 45
Proof of the Cosine Limit

Proof.


                 1 − cos θ   1 − cos θ 1 + cos θ     1 − cos2 θ
                           =          ·           =
                     θ           θ      1 + cos θ   θ(1 + cos θ)
                                         sin2 θ      sin θ     sin θ
                                  =                =       ·
                                      θ(1 + cos θ)     θ     1 + cos θ

So
                                  (            ) (              )
                      1 − cos θ          sin θ          sin θ
                  lim           = lim           · lim
                  θ→0     θ         θ→0 θ         θ→0 1 + cos θ

                                = 1 · 0 = 0.


                                                                          .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)          Section 1.4 Calculating Limits           September 15, 2010   42 / 45
Try these


Example
        tan θ
 1. lim
      θ→0 θ
        sin 2θ
 2. lim
    θ→0    θ




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   43 / 45
Try these


Example
        tan θ
 1. lim
      θ→0 θ
        sin 2θ
 2. lim
    θ→0    θ


Answer

 1. 1
 2. 2



                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   43 / 45
Solutions

 1. Use the basic trigonometric limit and the definition of tangent.

               tan θ        sin θ        sin θ         1        1
             lim     = lim         = lim       · lim       = 1 · = 1.
            θ→0 θ      θ→0 θ cos θ   θ→0 θ       θ→0 cos θ      1




                                                                   .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   44 / 45
Solutions

 1. Use the basic trigonometric limit and the definition of tangent.

               tan θ        sin θ        sin θ         1        1
             lim     = lim         = lim       · lim       = 1 · = 1.
            θ→0 θ      θ→0 θ cos θ   θ→0 θ       θ→0 cos θ      1


 2. Change the variable:

                         sin 2θ       sin 2θ           sin 2θ
                   lim          = lim        = 2 · lim        =2·1=2
                 θ→0        θ    2θ→0 2θ · 1      2θ→0 2θ
                                           2




                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   44 / 45
Solutions

 1. Use the basic trigonometric limit and the definition of tangent.

               tan θ        sin θ        sin θ         1        1
             lim     = lim         = lim       · lim       = 1 · = 1.
            θ→0 θ      θ→0 θ cos θ   θ→0 θ       θ→0 cos θ      1


 2. Change the variable:

                         sin 2θ       sin 2θ           sin 2θ
                   lim          = lim        = 2 · lim        =2·1=2
                 θ→0        θ    2θ→0 2θ · 1      2θ→0 2θ
                                           2

      OR use a trigonometric identity:

          sin 2θ       2 sin θ cos θ          sin θ
      lim        = lim               = 2· lim       · lim cos θ = 2·1·1 = 2
      θ→0    θ     θ→0       θ           θ→0 θ       θ→0


                                                                    .   .     .      .      .    .

 V63.0121.041, Calculus I (NYU)    Section 1.4 Calculating Limits           September 15, 2010   44 / 45
Summary




     The limit laws allow us to compute limits reasonably.
     BUT we cannot make up extra laws otherwise we get into trouble.




                                                                  .   .     .      .      .    .

V63.0121.041, Calculus I (NYU)   Section 1.4 Calculating Limits           September 15, 2010   45 / 45

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Lesson 4: Calculating Limits (Section 41 slides)

  • 1. Section 1.4 Calculating Limits V63.0121.041, Calculus I New York University September 15, 2010 Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper . . . . . .
  • 2. Announcements First written homework due today (put it in the envelope) Remember to put your lecture and recitation section numbers on your paper . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 2 / 45
  • 3. Yoda on teaching a concepts course “You must unlearn what you have learned.” In other words, we are building up concepts and allowing ourselves only to speak in terms of what we personally have produced. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 3 / 45
  • 4. Objectives Know basic limits like lim x = a and lim c = c. x→a x→a Use the limit laws to compute elementary limits. Use algebra to simplify limits. Understand and state the Squeeze Theorem. Use the Squeeze Theorem to demonstrate a limit. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 4 / 45
  • 5. Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 5 / 45
  • 6. Heuristic Definition of a Limit Definition We write lim f(x) = L x→a and say “the limit of f(x), as x approaches a, equals L” if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 6 / 45
  • 7. The error-tolerance game A game between two players (Dana and Emerson) to decide if a limit lim f(x) exists. x→a Step 1 Dana proposes L to be the limit. Step 2 Emerson challenges with an “error” level around L. Step 3 Dana chooses a “tolerance” level around a so that points x within that tolerance of a (not counting a itself) are taken to values y within the error level of L. If Dana cannot, Emerson wins and the limit cannot be L. Step 4 If Dana’s move is a good one, Emerson can challenge again or give up. If Emerson gives up, Dana wins and the limit is L. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 7 / 45
  • 8. The error-tolerance game L . . a . To be legit, the part of the graph inside the blue (vertical) strip must also be inside the green (horizontal) strip. If Emerson shrinks the error, Dana can still win. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 8 / 45
  • 9. Limit FAIL: Jump y . . . 1 . x . . Part of graph . 1. − inside blue is not inside green . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
  • 10. Limit FAIL: Jump y . . Part of graph inside blue is not . . 1 inside green . x . . 1. − . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
  • 11. Limit FAIL: Jump y . . Part of graph inside blue is not . . 1 inside green . x . . 1. − |x| So lim does not exist. x→0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 9 / 45
  • 12. Limit FAIL: unboundedness y . . 1 lim+ does not exist x→0 x because the function is unbounded near 0 .?. L . x . 0 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 10 / 45
  • 13. Limit EPIC FAIL (π ) Here is a graph of the function f(x) = sin : x y . . . 1 . x . . 1. − For every y in [−1, 1], there are infinitely many points x arbitrarily close to zero where f(x) = y. So lim f(x) cannot exist. x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 11 / 45
  • 14. Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 12 / 45
  • 15. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 13 / 45
  • 16. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The first is tautological, the second is trivial. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 13 / 45
  • 17. ET game for f(x) = x y . . x . . . . . . .
  • 18. ET game for f(x) = x y . . x . . . . . . .
  • 19. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 20. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 21. ET game for f(x) = x y . . . a . . x . a . . . . . . .
  • 22. ET game for f(x) = x y . . . a . . x . a . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 14 / 45
  • 23. ET game for f(x) = x y . . . a . . x . a . Setting error equal to tolerance works! . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 14 / 45
  • 24. ET game for f(x) = c . . . . . . .
  • 25. ET game for f(x) = c y . . x . . . . . . .
  • 26. ET game for f(x) = c y . . x . . . . . . .
  • 27. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 28. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 29. ET game for f(x) = c y . . c . . . x . a . . . . . . .
  • 30. ET game for f(x) = c y . . c . . . x . a . any tolerance works! . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 15 / 45
  • 31. Really basic limits Fact Let c be a constant and a a real number. (i) lim x = a x→a (ii) lim c = c x→a Proof. The first is tautological, the second is trivial. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 16 / 45
  • 32. Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 17 / 45
  • 33. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 18 / 45
  • 34. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 18 / 45
  • 35. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
  • 36. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
  • 37. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL (error scales) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 19 / 45
  • 38. Justification of the scaling law errors scale: If f(x) is e away from L, then (c · f(x) − c · L) = c · (f(x) − L) = c · e That is, (c · f)(x) is c · e away from cL, So if Emerson gives us an error of 1 (for instance), Dana can use the fact that lim f(x) = L to find a tolerance for f and g x→a corresponding to the error 1/c. Dana wins the round. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 20 / 45
  • 39. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 21 / 45
  • 40. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a 4. lim [f(x)g(x)] = L · M x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 22 / 45
  • 41. Limits and arithmetic Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M (errors add) x→a 2. lim [f(x) − g(x)] = L − M (combination of adding and scaling) x→a 3. lim [cf(x)] = cL (error scales) x→a 4. lim [f(x)g(x)] = L · M (more complicated, but doable) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 22 / 45
  • 42. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. x→a g(x) M . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 23 / 45
  • 43. Caution! The quotient rule for limits says that if lim g(x) ̸= 0, then x→a f(x) limx→a f(x) lim = x→a g(x) limx→a g(x) It does NOT say that if lim g(x) = 0, then x→a f(x) lim does not exist x→a g(x) In fact, limits of quotients where numerator and denominator both tend to 0 are exactly where the magic happens. more about this later . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 24 / 45
  • 44. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) x→a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 45. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 46. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 47. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a x→a √ √ 8. lim n x = n a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 48. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a (follows from 6) x→a √ √ 8. lim n x = n a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 49. Limits and arithmetic II Fact (Continued) f(x) L 5. lim = , if M ̸= 0. g(x) x→a M [ ]n n 6. lim [f(x)] = lim f(x) (follows from 4 repeatedly) x→a x→a n n 7. lim x = a (follows from 6) x→a √ √ 8. lim n x = n a x→a √ √ 9. lim n f(x) = n lim f(x) (If n is even, we must additionally assume x→a x→a that lim f(x) > 0) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 25 / 45
  • 50. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 51. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) lim x2 + 2x + 4 x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 52. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 53. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 54. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 = (3)2 + 2 · 3 + 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 55. Applying the limit laws Example ( ) Find lim x2 + 2x + 4 . x→3 Solution By applying the limit laws repeatedly: ( ) ( ) lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4) x→3 x→3 x→3 x→3 ( )2 = lim x + 2 · lim (x) + 4 x→3 x→3 = (3)2 + 2 · 3 + 4 = 9 + 6 + 4 = 19. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 26 / 45
  • 56. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 27 / 45
  • 57. Your turn Example x2 + 2x + 4 Find lim x→3 x3 + 11 Solution 19 1 The answer is = . 38 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 27 / 45
  • 58. Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f, then lim f(x) = f(a) x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 28 / 45
  • 59. Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 29 / 45
  • 60. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
  • 61. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
  • 62. Limits do not see the point! (in a good way) Theorem If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L. x→a x→a Example x2 + 2x + 1 Find lim , if it exists. x→−1 x+1 Solution x2 + 2x + 1 Since = x + 1 whenever x ̸= −1, and since x+1 x2 + 2x + 1 lim x + 1 = 0, we have lim = 0. x→−1 x→−1 x+1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 30 / 45
  • 63. x2 + 2x + 1 ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . .
  • 64. x2 + 2x + 1 ET game for f(x) = x+1 y . . . x . − . 1 Even if f(−1) were something else, it would not effect the limit. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 31 / 45
  • 65. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 66. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 67. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 68. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 69. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 70. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 71. Limit of a function defined piecewise at a boundary point Example Let { x2 x ≥ 0 . f(x) = −x x < 0 Does lim f(x) exist? x→0 Solution We have MTP DSP lim+ f(x) = lim+ x2 = 02 = 0 x→0 x→0 Likewise: lim f(x) = lim −x = −0 = 0 x→0− x→0− . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45
  • 72. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
  • 73. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
  • 74. Finding limits by algebraic manipulations Example √ x−2 Find lim . x→4 x − 4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x−2 x−2 lim = lim √ √ x→4 x − 4 x→4 ( x − 2)( x + 2) 1 1 = lim √ = x→4 x+2 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 33 / 45
  • 75. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 Find lim f(x) if it exists. x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 76. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 Find lim f(x) if it exists. x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 77. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 78. Your turn Example Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 79. Your turn Example . Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 80. Your turn Example . Let { 1 − x2 x≥1 f(x) = 2x x<1 . . Find lim f(x) if it exists. 1 . x→1 Solution We have ( ) DSP lim+ f(x) = lim+ 1 − x2 = 0 x→1 x→1 DSP lim f(x) = lim (2x) = 2 − − x→1 x→1 The left- and right-hand limits disagree, so the limit does not exist. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45
  • 81. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
  • 82. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a Too many verbs . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
  • 83. A message from the Mathematical Grammar Police Please do not say “ lim f(x) = DNE.” Does not compute. x→a Too many verbs Leads to FALSE limit laws like “If lim f(x) DNE and lim g(x) DNE, x→a x→a then lim (f(x) + g(x)) DNE.” x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 35 / 45
  • 84. Two More Important Limit Theorems Theorem If f(x) ≤ g(x) when x is near a (except possibly at a), then lim f(x) ≤ lim g(x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f(x) = lim h(x) = L, x→a x→a then lim g(x) = L. x→a . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 36 / 45
  • 85. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
  • 86. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) Show that lim x2 sin = 0. x→0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
  • 87. Using the Squeeze Theorem We can use the Squeeze Theorem to replace complicated expressions with simple ones when taking the limit. Example (π ) Show that lim x2 sin = 0. x→0 x Solution We have for all x, (π ) (π ) −1 ≤ sin ≤ 1 =⇒ −x2 ≤ x2 sin ≤ x2 x x The left and right sides go to zero as x → 0. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 37 / 45
  • 88. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
  • 89. Illustration of the Squeeze Theorem y . . (x) = x2 h . x . .(x) = −x2 f . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
  • 90. Illustration of the Squeeze Theorem y . . (x) = x2 h (π ) . (x) = x2 sin g x . x . .(x) = −x2 f . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 38 / 45
  • 91. Outline Recall: The concept of limit Basic Limits Limit Laws The direct substitution property Limits with Algebra Two more limit theorems Two important trigonometric limits . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 39 / 45
  • 92. Two important trigonometric limits Theorem The following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 40 / 45
  • 93. Proof of the Sine Limit Proof. Notice θ . θ . θ . 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 94. Proof of the Sine Limit Proof. Notice sin θ ≤ θ . in θ . s θ . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 95. Proof of the Sine Limit Proof. Notice sin θ ≤ θ tan θ . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 96. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 97. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ . c . os θ 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 98. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ Take reciprocals: . c . os θ 1 . sin θ 1≥ ≥ cos θ θ . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 99. Proof of the Sine Limit Proof. Notice θ sin θ ≤ θ ≤ 2 tan ≤ tan θ 2 Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ . in θ . .an θ s θ t . θ Take reciprocals: . c . os θ 1 . sin θ 1≥ ≥ cos θ θ As θ → 0, the left and right sides tend to 1. So, then, must the middle expression. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 41 / 45
  • 100. Proof of the Cosine Limit Proof. 1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ = · = θ θ 1 + cos θ θ(1 + cos θ) sin2 θ sin θ sin θ = = · θ(1 + cos θ) θ 1 + cos θ So ( ) ( ) 1 − cos θ sin θ sin θ lim = lim · lim θ→0 θ θ→0 θ θ→0 1 + cos θ = 1 · 0 = 0. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 42 / 45
  • 101. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 43 / 45
  • 102. Try these Example tan θ 1. lim θ→0 θ sin 2θ 2. lim θ→0 θ Answer 1. 1 2. 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 43 / 45
  • 103. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
  • 104. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
  • 105. Solutions 1. Use the basic trigonometric limit and the definition of tangent. tan θ sin θ sin θ 1 1 lim = lim = lim · lim = 1 · = 1. θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1 2. Change the variable: sin 2θ sin 2θ sin 2θ lim = lim = 2 · lim =2·1=2 θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ 2 OR use a trigonometric identity: sin 2θ 2 sin θ cos θ sin θ lim = lim = 2· lim · lim cos θ = 2·1·1 = 2 θ→0 θ θ→0 θ θ→0 θ θ→0 . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 44 / 45
  • 106. Summary The limit laws allow us to compute limits reasonably. BUT we cannot make up extra laws otherwise we get into trouble. . . . . . . V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 45 / 45