L’Hôpital’s Rule Chapter 9.2

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

Indeterminate Form 0 0

Indeterminate Form 0 0 The graph seems to suggest that lim 𝑥→ 4 + 𝑥−4 𝑥−4 =0 If so, then 0 0 does not mean that the limit does not exist In fact, 0 0 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

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

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

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

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 ?

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

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.

L’Hôpital’s Rule The requirement that 𝑓 𝑎 =𝑔 𝑎 =0 establishes that this works for the 0 0 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

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

Example 1

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

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→0 1 2 1+𝑥 1 = lim 𝑥→0 1 2 1+𝑥 = 1 2 1+0 = 1 2

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

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

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

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

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

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 in this case does not exist (or approaches +∞)

Example 3: One-sided Limits

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

Example 3: One-sided Limits

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 0 0 or ∞ ∞

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

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

Example 4

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

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

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 ∞ ∞

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

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

Example 5

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

Example 6

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 0 0 or ∞ ∞ . We can often rewrite the expression to obtain one or the other form. In this case lim 𝑥→∞ sin 1 𝑥 1 𝑥

Example 6: Indeterminate Form ∞⋅0 lim 𝑥→∞ 𝑥⋅ sin 1 𝑥 lim 𝑥→∞ sin 1 𝑥 1 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

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

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

Example 7: ∞−∞

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

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

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

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

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

Example 8: The Indeterminate Form 1 ∞

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

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

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

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 ⋅ 1 1+ 1 𝑥 − 1 𝑥 2 = lim 𝑥→∞ 1 1+ 1 𝑥 =1

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

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

Example 9: Working with 0 0

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

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

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

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

Example 10: Working with ∞ 0

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

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