1. 12.2 Limit Theorems
Proverbs 1:7. “The fear of the Lord is the beginning of
knowledge, but fools despise wisdom and instruction.”

2. If f(x) is equal to a constant, k,
then lim f ( x ) = k
x→c

3. If f(x) is equal to a constant, k,
then lim f ( x ) = k
x→c
The limit of a constant is that constant.

4. If f(x) is equal to a constant, k,
then lim f ( x ) = k
x→c
The limit of a constant is that constant.
If f ( x ) = 4, evaluate lim f ( x )
x→1

5. If f(x) is equal to a constant, k,
then lim f ( x ) = k
x→c
The limit of a constant is that constant.
If f ( x ) = 4, evaluate lim f ( x )
x→1
4

6. If f(x) is equal to a constant, k,
then lim f ( x ) = k
x→c
The limit of a constant is that constant.
If f ( x ) = 4, evaluate lim f ( x )
x→1
4
(sketch and show graphically)

7. m
If f ( x ) = x , m ∈+° ,
m m
then lim x = c
x→c

8. m
If f ( x ) = x , m ∈+° ,
m m
then lim x = c
x→c
Always try to evaluate limits by using
substitution first!

9. m
If f ( x ) = x , m ∈+° ,
m m
then lim x = c
x→c
Always try to evaluate limits by using
substitution first!
3 3
lim x = 2 = 8
x→2

10. m
If f ( x ) = x , m ∈+° ,
m m
then lim x = c
x→c
Always try to evaluate limits by using
substitution first!
3 3
lim x = 2 = 8
x→2
(sketch and show graphically)

11. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c

12. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c
1. lim ⎡ f ( x ) + g ( x ) ⎤ = lim f ( x ) + lim g ( x )
⎣ ⎦ x→c
x→c x→c

13. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c
1. lim ⎡ f ( x ) + g ( x ) ⎤ = lim f ( x ) + lim g ( x )
⎣ ⎦ x→c
x→c x→c
The limit of the sum is the sum of the limits

14. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c
1. lim ⎡ f ( x ) + g ( x ) ⎤ = lim f ( x ) + lim g ( x )
⎣ ⎦ x→c
x→c x→c
The limit of the sum is the sum of the limits
2. lim ⎡ f ( x ) − g ( x ) ⎤ = lim f ( x ) − lim g ( x )
⎣ ⎦ x→c
x→c x→c

15. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c
1. lim ⎡ f ( x ) + g ( x ) ⎤ = lim f ( x ) + lim g ( x )
⎣ ⎦ x→c
x→c x→c
The limit of the sum is the sum of the limits
2. lim ⎡ f ( x ) − g ( x ) ⎤ = lim f ( x ) − lim g ( x )
⎣ ⎦ x→c
x→c x→c
3. lim ⎡ f ( x ) ⋅ g ( x ) ⎤ = lim f ( x ) ⋅ lim g ( x )
⎣ ⎦ x→c
x→c x→c

16. If lim f ( x ) = L and lim g ( x ) = M both exist, then
x→c x→c
1. lim ⎡ f ( x ) + g ( x ) ⎤ = lim f ( x ) + lim g ( x )
⎣ ⎦ x→c
x→c x→c
The limit of the sum is the sum of the limits
2. lim ⎡ f ( x ) − g ( x ) ⎤ = lim f ( x ) − lim g ( x )
⎣ ⎦ x→c
x→c x→c
3. lim ⎡ f ( x ) ⋅ g ( x ) ⎤ = lim f ( x ) ⋅ lim g ( x )
⎣ ⎦ x→c
x→c x→c
f ( x ) lim f ( x )
4. lim = x→c
, g ( x ) & lim g ( x ) ≠ 0
x→c g ( x ) lim g ( x ) x→c
x→c

17. Find each limit:

18. Find each limit:
1. lim 10
x→3

19. Find each limit:
1. lim 10
x→3
10

20. Find each limit:
1. lim 10 2. lim x 3
x→3 x→−2
10

21. Find each limit:
1. lim 10 2. lim x 3
x→3 x→−2
10 ( −2 ) 3
−8

22. Find each limit:
1. lim 10 2. lim x 3
x→3 x→−2
10 ( −2 ) 3
−8
2
3. lim x + 5x
x→3

23. Find each limit:
1. lim 10 2. lim x 3
x→3 x→−2
10 ( −2 ) 3
−8
2
3. lim x + 5x
x→3
2
( 3) + 5 ( 3)
24

24. Find each limit:
3
x − 4x
4. lim 2
x→−1 x + x

25. Find each limit:
3
x − 4x Rats! Can’t use substitution as
4. lim 2
x→−1 x + x we get zero in the denominator.

26. Find each limit:
3
x − 4x Rats! Can’t use substitution as
4. lim 2
x→−1 x + x we get zero in the denominator.
x(x − 4)
2
lim
x→−1 x ( x + 1)

27. Find each limit:
3
x − 4x Rats! Can’t use substitution as
4. lim 2
x→−1 x + x we get zero in the denominator.
x(x − 4)
2
lim
x→−1 x ( x + 1)
2
x −4
lim
x→−1 x + 1

28. Find each limit:
3
x − 4x Rats! Can’t use substitution as
4. lim 2
x→−1 x + x we get zero in the denominator.
x(x − 4)
2
lim
x→−1 x ( x + 1)
x −42 Nice try ... but still have zero in
lim the denominator. Check out the
x→−1 x + 1
graph of the original function.

29. Find each limit:
3
x − 4x Rats! Can’t use substitution as
4. lim 2
x→−1 x + x we get zero in the denominator.
x(x − 4)
2
lim
x→−1 x ( x + 1)
x −42 Nice try ... but still have zero in
lim the denominator. Check out the
x→−1 x + 1
graph of the original function.
The limit does not exist!

30. Find each limit:
2
x − 36
5. lim
x→−6 x + 6

31. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.

32. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.
lim
( x − 6 )( x + 6 )
x→−6 ( x + 6)

33. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.
lim
( x − 6 )( x + 6 )
x→−6 ( x + 6)
lim ( x − 6 )
x→−6

34. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.
lim
( x − 6 )( x + 6 )
x→−6 ( x + 6)
lim ( x − 6 )
x→−6
−12

35. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.
lim
( x − 6 )( x + 6 )
x→−6 ( x + 6)
lim ( x − 6 )
x→−6
−12 Verify graphically

36. Find each limit:
2
x − 36 Again ... can’t use substitution.
5. lim
x→−6 x + 6 Let’s try factoring again.
lim
( x − 6 )( x + 6 )
x→−6 ( x + 6)
lim ( x − 6 )
x→−6
−12 Verify graphically
If substitution can’t be used, try to manipulate the
function until substitution will work.

37. Find each limit:
x−3
6. lim 2
x→3 x − 9

38. Find each limit:
x−3
6. lim 2
x→3 x − 9
x−3
lim
x→3 ( x + 3) ( x − 3)

39. Find each limit:
x−3
6. lim 2
x→3 x − 9
x−3
lim
x→3 ( x + 3) ( x − 3)
1
6

40. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.

41. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x
7. lim
x→∞ 2x 2 − 5

42. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x Substitution yields infinity over
7. lim infinity which is indeterminable.
x→∞ 2x 2 − 5

43. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x Substitution yields infinity over
7. lim infinity which is indeterminable.
x→∞ 2x 2 − 5
1
2
Multiply by: x
1
2
x

44. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x Substitution yields infinity over
7. lim infinity which is indeterminable.
x→∞ 2x 2 − 5
1
2
13 Multiply by: x
1+ 1
lim x
x→∞ 5 x 2
2− 2
x

45. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x Substitution yields infinity over
7. lim infinity which is indeterminable.
x→∞ 2x 2 − 5
1
2
13 Multiply by: x
1+ 1
lim x
x→∞ 5 x 2
2− 2
x
k
Recall: lim =0
x→∞ x

46. Limits approaching infinity require their own
unique techniques. Let’s get an introduction
to these techniques.
2
x + 13x Substitution yields infinity over
7. lim infinity which is indeterminable.
x→∞ 2x 2 − 5
1
2
13 Multiply by: x
1+ 1
lim x
x→∞ 5 x 2
2− 2
x
k
1 Recall: lim =0
x→∞ x
2

47. 2
7x − 19
8. lim 4
x→−∞ x + 8

48. 2
1
7x − 19 x 4
8. lim 4 Multiply by:
x→−∞ x + 8 1
4
x

49. 2
1
7x − 19 x 4
8. lim 4 Multiply by:
x→−∞ x + 8 1
4
x
7 19
2
− 4
lim x x
x→−∞ 8
1+ 4
x

50. 2
1
7x − 19 x 4
8. lim 4 Multiply by:
x→−∞ x + 8 1
4
x
7 19
2
− 4
lim x x
x→−∞ 8
1+ 4
x
0
1

51. 2
1
7x − 19 x 4
8. lim 4 Multiply by:
x→−∞ x + 8 1
4
x
7 19
2
− 4
lim x x
x→−∞ 8
1+ 4
x
0
1
0

52. 2
1
7x − 19 x 4
8. lim 4 Multiply by:
x→−∞ x + 8 1
4
x
7 19
2
− 4
lim x x
x→−∞ 8
1+ 4
x
0
1
0 We can verify graphically.

53. HW #2
The only way of finding the limits of the possible is by going
beyond them into the impossible.
Arthur C. Clarke