Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 25

Answer

The solution is $$\lim_{x\to\pi}\frac{\cos x+1}{(x-\pi)^2}=\frac{1}{2}.$$

Work Step by Step

We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to\pi}\frac{\cos x+1}{(x-\pi)^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\pi}\frac{(\cos x+1)'}{((x-\pi)^2)'}=\lim_{x\to\pi}\frac{-\sin x}{2(x-\pi)}=\left[\frac{-\sin \pi}{2(\pi-\pi)}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\pi}\frac{(-\sin x)'}{(2x-2\pi)'}=\lim_{x\to\pi}\frac{-\cos x}{2}=\frac{-\cos\pi}{2}=\frac{1}{2}.$$
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