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Chapter 4 | Applications of Derivatives
−∞, or the limit is one-sided.
Example 4.39 Applying L’Hôpital’s Rule (∞/∞ Case)
Evaluate each of the following limits by applying L’Hôpital’s rule. a. lim x →∞ 3 x +5 2 x +1
ln x cot x
lim x →0 +
b.
Solution a. Since 3 x +5 and 2 x +1 are first-degree polynomials with positive leading coefficients, lim x →∞ (3 x +5) =∞ and lim x →∞ (2 x +1) =∞. Therefore, we apply L’Hôpital’s rule and obtain lim x →∞ 3 x +5 2 x +1 / x = lim x →∞ 3 2 = 3 2 . Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of x in the denominator. In doing so, we saw that lim x →∞ 3 x +5 2 x +1 = lim x →∞ 3+5/ x 2 x +1/ x = 3 2 .
L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit. b. Here, lim x →0 + ln x =−∞ and lim x →0 +
cot x =∞. Therefore, we can apply L’Hôpital’s rule and obtain
1/ x −csc 2 x
ln x cot x = lim
1 − x csc 2 x .
lim x →0 +
= lim
x →0 +
x →0 +
Now as x →0 + , csc 2 x →∞. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of csc x to write lim x →0 + 1 − x csc 2 x = lim x →0 + sin 2 x − x .
sin 2 x =0 and lim x →0 +
Now lim
x =0, so we apply L’Hôpital’s rule again. We find
x →0 +
sin 2 x
0 −1 =0.
2sin x cos x
lim x →0 +
− x = lim
−1 =
x →0 +
We conclude that
ln x cot x =0.
lim x →0 +
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