Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 41



Work Step by Step

Here, we have $\lim\limits_{x \to 1} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 1} \dfrac{-1+1/x)}{1-(1/x)+\ln x}=\dfrac{0}{0}$ Now, again apply L-Hospital's rule. $\lim\limits_{x \to 1} \dfrac{-1/x^2}{1+x/x^2}=-\dfrac{1/1}{1+1/1}$ or, $-\dfrac{1}{1+1}=-\dfrac{1}{2}$
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