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: 20

Answer

$\dfrac{1}{1 +\pi}$

Work Step by Step

Consider: $\lim\limits_{x \to 1}f(x)=\lim\limits_{x \to 1}\dfrac{x-1}{\ln x-\sin (\pi x)}$ Need to check that the limit has an indeterminate form. Thus, $f(1)=\dfrac{1-1}{\ln 1-\sin (1) \pi}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to 1}\dfrac{1}{\dfrac{1}{x}-\pi \cos (\pi x)}=\dfrac{1}{\dfrac{1}{1}-\pi \cos (\pi)}=\dfrac{1}{1 +\pi}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.