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: 34



Work Step by Step

Here, we have $\lim\limits_{x \to 0^{+}} f(0)=\dfrac{0}{0}$ 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 0^{+}} \dfrac{e^x/e^x-1}{1/x}=\lim\limits_{x \to 0^{+}} \dfrac{xe^x}{e^x-1}=\dfrac{0}{0}$ Now, again apply L-Hospital's rule. $\lim\limits_{x \to 0^{+}} \dfrac{e^x+xe^x}{e^x}=1$
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.