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1202 ch 12 day 2

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1202 ch 12 day 2

  1. 1. 12.2 Limit TheoremsProverbs 1:7. “The fear of the Lord is the beginning ofknowledge, but fools despise wisdom and instruction.”
  2. 2. If f(x) is equal to a constant, k, then lim f ( x ) = k x→c
  3. 3. If f(x) is equal to a constant, k, then lim f ( x ) = k x→cThe limit of a constant is that constant.
  4. 4. If f(x) is equal to a constant, k, then lim f ( x ) = k x→cThe limit of a constant is that constant. If f ( x ) = 4, evaluate lim f ( x ) x→1
  5. 5. If f(x) is equal to a constant, k, then lim f ( x ) = k x→cThe limit of a constant is that constant. If f ( x ) = 4, evaluate lim f ( x ) x→1 4
  6. 6. If f(x) is equal to a constant, k, then lim f ( x ) = k x→cThe limit of a constant is that constant. If f ( x ) = 4, evaluate lim f ( x ) x→1 4 (sketch and show graphically)
  7. 7. mIf f ( x ) = x , m ∈+° , m m then lim x = c x→c
  8. 8. m If f ( x ) = x , m ∈+° , m m then lim x = c x→cAlways try to evaluate limits by usingsubstitution first!
  9. 9. m If f ( x ) = x , m ∈+° , m m then lim x = c x→cAlways try to evaluate limits by usingsubstitution first! 3 3 lim x = 2 = 8 x→2
  10. 10. m If f ( x ) = x , m ∈+° , m m then lim x = c x→cAlways try to evaluate limits by usingsubstitution first! 3 3 lim x = 2 = 8 x→2 (sketch and show graphically)
  11. 11. If lim f ( x ) = L and lim g ( x ) = M both exist, then x→c x→c
  12. 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. 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. 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. 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. 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. 17. Find each limit:
  18. 18. Find each limit: 1. lim 10 x→3
  19. 19. Find each limit: 1. lim 10 x→3 10
  20. 20. Find each limit: 1. lim 10 2. lim x 3 x→3 x→−2 10
  21. 21. Find each limit: 1. lim 10 2. lim x 3 x→3 x→−2 10 ( −2 ) 3 −8
  22. 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. 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. 24. Find each limit: 3 x − 4x 4. lim 2 x→−1 x + x
  25. 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. 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. 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. 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. 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. 30. Find each limit: 2 x − 36 5. lim x→−6 x + 6
  31. 31. Find each limit: 2 x − 36 Again ... can’t use substitution. 5. lim x→−6 x + 6 Let’s try factoring again.
  32. 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. 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. 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. 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. 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 graphicallyIf substitution can’t be used, try to manipulate thefunction until substitution will work.
  37. 37. Find each limit: x−3 6. lim 2 x→3 x − 9
  38. 38. Find each limit: x−3 6. lim 2 x→3 x − 9 x−3 lim x→3 ( x + 3) ( x − 3)
  39. 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. 40. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto these techniques.
  41. 41. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto these techniques. 2 x + 13x 7. lim x→∞ 2x 2 − 5
  42. 42. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto these techniques. 2 x + 13x Substitution yields infinity over 7. lim infinity which is indeterminable. x→∞ 2x 2 − 5
  43. 43. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto 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. 44. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto 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. 45. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto 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. 46. Limits approaching infinity require their ownunique techniques. Let’s get an introductionto 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. 47. 2 7x − 198. lim 4 x→−∞ x + 8
  48. 48. 2 1 7x − 19 x 48. lim 4 Multiply by: x→−∞ x + 8 1 4 x
  49. 49. 2 1 7x − 19 x 48. lim 4 Multiply by: x→−∞ x + 8 1 4 x 7 19 2 − 4 lim x x x→−∞ 8 1+ 4 x
  50. 50. 2 1 7x − 19 x 48. 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. 51. 2 1 7x − 19 x 48. 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. 52. 2 1 7x − 19 x 48. 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. 53. HW #2The only way of finding the limits of the possible is by goingbeyond them into the impossible. Arthur C. Clarke

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