Answer
0
Work Step by Step
The original limit is of an indeterminate form $\frac{0}{0}$. Hence, we use L'Hopital's:
$\displaystyle\lim_{x\to1}\frac{(\ln x)^2}{x-1}=\lim_{x\to1}\frac{2\ln x}{1}=\lim_{x\to1}2\ln x=\fbox{$0$}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 580: 73
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