## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 8

#### Answer

$-10$

#### Work Step by Step

Consider: $\lim\limits_{x \to -5}f(x)=\lim\limits_{x \to -5}\dfrac{x^2-25}{x +5}$ Now, $f(-5)=\dfrac{(-5)^2 -25}{-5+5}=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Thus: $\lim\limits_{x \to -5}\dfrac{2x}{1}=2(-5)=-10$

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.