1. L’Hôpital’s Rule
Chapter 9.2

2. Indeterminate Form 0 0 What is lim 𝑥→ 4 + 𝑥−4 𝑥−4 ?
Substitution yields 0 0
Does this mean that the limit does not exist?
Lets take a look at the graph of 𝑓 𝑥 = 𝑥−4 𝑥−4

3. Indeterminate Form 0 0

4. Indeterminate Form 0 0
The graph seems to suggest that lim 𝑥→ 𝑥−4 𝑥−4 =0
If so, then does not mean that the limit does not exist
In fact, is one of several indeterminate forms
These are expressions that, when they appear as the result of taking a limit, do not allow us to determine whether the limit exists or, if it does, what its value is

5. Indeterminate Form 0 0
We can verify that the limit in the previous expression is zero in the following way:
lim 𝑥→ 𝑥−4 𝑥−4 = lim 𝑥→ 𝑥−4 𝑥−4 ⋅ 𝑥−4 𝑥−4 =
lim 𝑥→ 𝑥−4 𝑥−4 𝑥−4 = lim 𝑥→ 𝑥−4 = 4−4 =0

6. Indeterminate Form 0 0
We have seen the indeterminate form in the following type of limit:
lim 𝑥→3 𝑥 2 −9 𝑥−3
Substituting yields 0 0 ; but if we simplify, then pass to the limit we get
lim 𝑥→3 𝑥 2 −9 𝑥−3 = lim 𝑥→3 (𝑥+3)(𝑥−3) 𝑥−3
lim 𝑥→3 (𝑥+3) =3+3=6

7. Indeterminate Form 0 0
It also appears in the definition of the derivative:
𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ
Substituting ℎ results in 0 0

8. Indeterminate Form 0 0
In the previous two examples, we could get around the indeterminate form if we could simplify the expression in such a way as to eliminate the part causing zero to appear in the denominator
But could we do this for lim 𝑥→0 sin 𝑥 𝑥 ?

9. Indeterminate Form 0 0
We proved earlier using a geometric argument that lim 𝑥→0 sin 𝑥 𝑥 =1
Can we use our knowledge of calculus to find a more general way to find limits such as these (if they exist)?
There is: the theorem is known as L’Hôpital’s Rule and applies when we get indeterminate forms while taking a limit
There are several indeterminate forms, not all of which are expressed as ratios, and we will learn how to handle them using this theorem

10. L’Hôpital’s Rule
THEOREM Suppose 𝑓 𝑎 =𝑔 𝑎 =0 and that 𝑓 and 𝑔 are differentiable on an open interval 𝐼 containing 𝑎, and that 𝑔 ′ 𝑥 ≠0 on 𝐼 if 𝑥≠𝑎. Then lim 𝑥→𝑎 𝑓 𝑥 𝑔 𝑥 = lim 𝑥→𝑎 𝑓 ′ 𝑥 𝑔 ′ 𝑥 if the latter limit exists.

11. L’Hôpital’s Rule
The requirement that 𝑓 𝑎 =𝑔 𝑎 =0 establishes that this works for the indeterminate form
Both numerator and denominator must be differentiable for values near 𝑎
The derivative of the denominator should not be zero at values near (but not necessarily equal to ) 𝑎
Also, most importantly, this is not the Quotient Rule; the theorem works by taking the derivatives of the numerator and denominator separately

12. Example 1
Estimate the limit graphically and then use L’Hôpital’s Rule to find the limit lim 𝑥→0 1+𝑥 −1 𝑥

13. Example 1

14. Example 1
We first note that substitution yields the 0 0 indeterminate form. If the form is not indeterminate, this theorem does not apply! The theorem asserts that lim 𝑥→0 1+𝑥 −1 𝑥 = lim 𝑥→0 𝑑 𝑑𝑥 1+𝑥 −1 𝑑 𝑑𝑥 𝑥

15. Example 1
The theorem asserts that lim 𝑥→0 1+𝑥 −1 𝑥 = lim 𝑥→0 𝑑 𝑑𝑥 1+𝑥 −1 𝑑 𝑑𝑥 𝑥 = lim 𝑥→ 𝑥 1 = lim 𝑥→ 𝑥 = = 1 2

16. Multiple Uses of the Theorem
In some cases, a first application of the theorem produces a form that again results in an indeterminate form
In that case, we can apply the theorem again to the resulting function
That is, if lim 𝑥→𝑎 𝑓 𝑥 𝑔 𝑥 = lim 𝑥→𝑎 𝑓 ′ 𝑥 𝑔 ′ 𝑥 = 0 0
then
lim 𝑥→𝑎 𝑓 𝑥 𝑔 𝑥 = lim 𝑥→𝑎 𝑓 ′′ 𝑥 𝑔 ′′ 𝑥

17. Example 2: Multiple Uses of the Theorem
Evaluate lim 𝑥→0 1+𝑥 −1− 𝑥 2 𝑥 2

18. Example 2: Multiple Uses of the Theorem
lim 𝑥→0 1+𝑥 −1− 𝑥 2 𝑥 2 = lim 𝑥→ 𝑥 − 1 2 2𝑥 = 0 0 Apply the theorem to the resulting function

19. Example 2: Multiple Uses of the Theorem
= lim 𝑥→ 𝑥 − 1 2 2𝑥 = 0 0 Apply the theorem to the resulting function lim 𝑥→0 −1 8 1+𝑥 3 2 =− 1 8

20. Example 3: One-sided Limits
Evaluate the following using L’Hôpital’s Rule:
lim 𝑥→ sin 𝑥 𝑥 2
lim 𝑥→ 0 − sin 𝑥 𝑥 2

21. Example 3: One-sided Limits
lim 𝑥→ sin 𝑥 𝑥 2
Apply the theorem
lim 𝑥→ sin 𝑥 𝑥 2 = lim 𝑥→ cos 𝑥 2𝑥 = 1 0
The limit in this case does not exist (or approaches +∞)

22. Example 3: One-sided Limits

23. Example 3: One-sided Limits
lim 𝑥→ 0 − sin 𝑥 𝑥 2
Apply the theorem
lim 𝑥→ 0 − sin 𝑥 𝑥 2 = lim 𝑥→ 0 − cos 𝑥 2𝑥 =− 1 0
The limit does not exist, or approaches −∞
Note that we must deduce that it approaches −∞ from the fact that the denominator has only negative values if 𝑥 approaches from the left

24. Example 3: One-sided Limits

25. Other Indeterminate Forms
With a slight modification in the proof, L’Hôpital’s Rule can also be applied to other indeterminate forms
Some of these are ∞ ∞ , ∞⋅0, ∞−∞
We can apply the theorem directly if the indeterminate form is ∞ ∞
For other forms, we must rewrite the function so that the indeterminate form is either or ∞ ∞

26. Example 4
Identify the indeterminate form and evaluate the limit using L’Hôpital’s Rule. Support your answer graphically. lim 𝑥→ 𝜋 2 sec 𝑥 1+ tan 𝑥

27. Example 4
lim 𝑥→ 𝜋 2 sec 𝑥 1+ tan 𝑥 By substituting we get ∞ ∞ We may apply the theorem: = lim 𝑥→ 𝜋 2 sec 𝑥 tan 𝑥 sec 2 𝑥 = lim 𝑥→ 𝜋 2 sin 𝑥 =1

28. Example 4

29. Limit at Infinity
If an indeterminate form results from taking the limit as 𝑥→∞ or as 𝑥→−∞, the theorem can be used to find the limit (if it exists)
The next example shows how to do this

30. Example 5
Identify the indeterminate form and evaluate the limit using L’Hôpital’s Rule. Support your answer graphically. lim 𝑥→∞ ln 𝑥 2 𝑥

31. Example 5
lim 𝑥→∞ ln 𝑥 2 𝑥 Since ln 𝑥 continues increasing (albeit very slowly) as 𝑥 increases, and the same for 𝑥 , then the limit above is equivalent to the indeterminate form ∞ ∞

32. Example 5
We can apply the theorem: lim 𝑥→∞ ln 𝑥 2 𝑥 = lim 𝑥→∞ 1 𝑥 1 𝑥 = lim 𝑥→∞ 𝑥 𝑥 But this again gives ∞ ∞ , so apply the theorem to the new function

33. Example 5
lim 𝑥→∞ 𝑥 𝑥 = lim 𝑥→∞ 1 2 𝑥 =0

34. Example 5

35. Example 6: Indeterminate Form ∞⋅0
Find
lim 𝑥→∞ 𝑥⋅ sin 1 𝑥
lim 𝑥→−∞ 𝑥⋅sin 1 𝑥

36. Example 6

37. Example 6: Indeterminate Form ∞⋅0
lim 𝑥→∞ 𝑥⋅ sin 1 𝑥
Note that 𝑥→∞ and 1 𝑥 →0, so this represents the indeterminate form ∞⋅0. But to use the theorem the form must be either or ∞ ∞ . We can often rewrite the expression to obtain one or the other form. In this case
lim 𝑥→∞ sin 1 𝑥 𝑥

38. Example 6: Indeterminate Form ∞⋅0
lim 𝑥→∞ 𝑥⋅ sin 1 𝑥
lim 𝑥→∞ sin 1 𝑥 𝑥
The form is now 0 0 , so apply the theorem
= lim 𝑥→∞ − 1 𝑥 2 cos 1 𝑥 − 1 𝑥 2 = lim 𝑥→∞ cos 1 𝑥 = cos 0 =1

39. Example 6: Indeterminate Form ∞⋅0
Find
lim 𝑥→−∞ 𝑥⋅sin 1 𝑥
Similarly,
lim 𝑥→−∞ 𝑥⋅ sin 1 𝑥 = lim 𝑥→−∞ sin 1 𝑥 𝑥 = lim 𝑥→−∞ − 1 𝑥 2 cos 1 𝑥 − 1 𝑥 2 =1

40. Example 7: ∞−∞
Find lim 𝑥→1 1 ln 𝑥 − 1 𝑥−1

41. Example 7: ∞−∞

42. Example 7: ∞−∞
lim 𝑥→1 1 ln 𝑥 − 1 𝑥−1 Substituting results in the indeterminate form ∞−∞. As before, we must rewrite the function so that the indeterminate form is 0 0 or ∞ ∞ . First, find a common denominator = lim 𝑥→1 𝑥−1− ln 𝑥 ln 𝑥 𝑥−1

43. Example 7: ∞−∞
= lim 𝑥→1 𝑥−1− ln 𝑥 ln 𝑥 𝑥−1 This now has the indeterminate form Apply the theorem: = lim 𝑥→1 1− 1 𝑥 ln 𝑥 + 1 𝑥 𝑥−1 = 0 0

44. Example 7: ∞−∞
= lim 𝑥→1 1− 1 𝑥 ln 𝑥 + 1 𝑥 𝑥−1 = 0 0 Apply the theorem again = lim 𝑥→1 1 𝑥 2 1 𝑥 − 1 𝑥 2 𝑥−1 + 1 𝑥 = 1 2

45. Indeterminate Forms 1 ∞ , 0 0 , ∞ 0
The indeterminate forms 1 ∞ , 0 0 , ∞ 0 can be handled by taking the natural logarithm of both sides
After finding the limit using the theorem, the result is then the exponent for 𝑒

46. Example 8: The Indeterminate Form 1 ∞
Find lim 𝑥→∞ 1+ 1 𝑥 𝑥

47. Example 8: The Indeterminate Form 1 ∞

48. Example 8: The Indeterminate Form 1 ∞
lim 𝑥→∞ 1+ 1 𝑥 𝑥 If 𝑓 𝑥 = 1+ 1 𝑥 𝑥 Then ln 𝑓 𝑥 =𝑥⋅ ln 1+ 1 𝑥

49. Example 8: The Indeterminate Form 1 ∞
ln 𝑓 𝑥 =𝑥⋅ ln 1+ 1 𝑥 Now, lim 𝑥→∞ ln 𝑓 𝑥 = lim 𝑥→∞ 𝑥⋅ ln 1+ 1 𝑥 The right side now has form ∞⋅0

50. Example 8: The Indeterminate Form 1 ∞
lim 𝑥→∞ ln 𝑓 𝑥 = lim 𝑥→∞ 𝑥⋅ ln 1+ 1 𝑥 = lim 𝑥→∞ ln 1+ 1 𝑥 1 𝑥 This results in the form 0 0 , so we can apply the theorem

51. Example 8: The Indeterminate Form 1 ∞
= lim 𝑥→∞ ln 1+ 1 𝑥 1 𝑥 = lim 𝑥→∞ 𝑑/𝑑𝑥 ln 1+ 1 𝑥 𝑑/𝑑𝑥 1 𝑥 = lim 𝑥→∞ − 1 𝑥 2 ⋅ 𝑥 − 1 𝑥 2 = lim 𝑥→∞ 𝑥 =1

52. Example 8: The Indeterminate Form 1 ∞
Finally we reason as follows: lim 𝑥→∞ 1+ 1 𝑥 𝑥 = lim 𝑥→∞ 𝑓(𝑥) = lim 𝑥→∞ 𝑒 ln (𝑓(𝑥) = 𝑒 1 =𝑒 Recall that it is from the discrete form of this formula, 1+ 1 𝑛 𝑛 , where 𝑛=1, 2, 3, …, that we arrived at the formula 𝐴 𝑡 = 𝐴 0 𝑒 𝑟𝑡

53. Example 9: Working with 0 0
Determine whether lim 𝑥→ 𝑥 𝑥 exists and find its value if it does.

54. Example 9: Working with 0 0

55. Example 9: Working with 0 0
As before we can use the logarithm properties to rewrite the function as ln 𝑓 𝑥 =𝑥⋅ ln 𝑥 Now, lim 𝑥→ 0 + 𝑥⋅ ln 𝑥 yields the indeterminate form 0⋅−∞. Rewrite this as = lim 𝑥→ 0 + ln 𝑥 1 𝑥 The indeterminate form is now ∞ ∞ ; apply the theorem

56. Example 9: Working with 0 0
= lim 𝑥→ 0 + ln 𝑥 1 𝑥 The indeterminate form is now ∞ ∞ ; apply the theorem = lim 𝑥→ 𝑥 − 1 𝑥 2 = lim 𝑥→ 0 + − 𝑥 2 𝑥 = lim 𝑥→ 0 + (−𝑥) =0

57. Example 9: Working with 0 0
Now “undo” the natural logarithm by making 𝑒 the base to get lim 𝑥→ 0 + 𝑥 𝑥 = 𝑒 0 =1

58. Example 10: Working with ∞ 0
Find lim 𝑥→∞ 𝑥 1 𝑥

59. Example 10: Working with ∞ 0

60. Example 10: Working with ∞ 0
If 𝑓 𝑥 = 𝑥 1 𝑥 , then ln 𝑓 𝑥 = 1 𝑥 ln 𝑥 . Find lim 𝑥→∞ 1 𝑥 ln 𝑥 = lim 𝑥→∞ ln 𝑥 𝑥 This yields the indeterminate form ∞ ∞ . Apply the theorem lim 𝑥→∞ 1 𝑥 1 = lim 𝑥→∞ 1 𝑥 =0

61. Example 10: Working with ∞ 0
Now apply 𝑒 as a base and we get lim 𝑥→∞ 𝑥 1 𝑥 = 𝑒 0 =1