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: 39

Answer

The solution is $$\lim_{x\to \pi/2^-}\frac{\tan x}{\frac{3}{2x-\pi}}=-\frac{2}{3}.$$

Work Step by Step

To calculate this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule." $$\lim_{x\to \pi/2^-}\frac{\tan x}{\frac{3}{2x-\pi}}=\lim_{x\to \pi/2^-}\frac{\frac{2x-\pi}{3}}{\frac{1}{\tan x}}=\left[\frac{\frac{2\cdot\pi/2-\pi}{3}}{\frac{1}{\tan(\pi/2^-)}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\pi/2^-}\frac{\left(\frac{2}{3}x-\frac{\pi}{3}\right)'}{\left(\frac{1}{\tan x}\right)'}=\lim_{x\to\pi/2^-}\frac{\frac{2}{3}}{-\frac{1}{\tan^2x}\cdot(\tan x)'}=\lim_{x\to\pi/2^-}\frac{\frac{2}{3}}{-\frac{1}{\tan^2 x}\cdot\frac{1}{\cos^2 x}}=\lim_{x\to\pi/2^-}\frac{\frac{2}{3}}{-\frac{1}{\sin^2x}}=\lim_{x\to\pi/2^-}=\lim_{x\to\pi/2^-}\frac{-2\sin^2x}{3}=\frac{-2\sin^2(\pi/2^-)}{3}=-\frac{2}{3}.$$
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