Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Review - Exercises - Page 359: 13

Answer

$\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x}) = \frac{1}{2}$

Work Step by Step

$\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x})$ $= \lim\limits_{x \to 1^+} [\frac{x~ln~x}{(x-1)~ln~x}-\frac{x-1}{(x-1)~ln~x}]$ $= \lim\limits_{x \to 1^+} [\frac{x~ln~x-x+1}{(x-1)~ln~x}] = \frac{0}{0}$ We can apply L'Hospital's Rule: $\lim\limits_{x \to 1^+} [\frac{ln~x+x~(1/x)-1}{ln~x+(x-1)/x}]$ $= \lim\limits_{x \to 1^+} [\frac{ln~x+1-1}{ln~x+(x-1)/x}] = \frac{0}{0}$ We can apply L'Hospital's Rule: $= \lim\limits_{x \to 1^+} \frac{1/x}{(1/x)+(1/x^2)} = \frac{1}{1+1} = \frac{1}{2}$ Therefore: $\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x}) = \frac{1}{2}$
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