Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 7

Answer

-1

Work Step by Step

Numerator at $x=-2:\,\,\,x^{2}+3x+2=(-2)^{2}+3(-2)+2=0$ Denominator at $x=-2:\,\,\,x+2=-2+2=0$ The function has the indeterminate form $\frac{0}{0}$ at $x=-2$. Transforming algebraically and canceling the common factor, we have $\frac{x^{2}+3x+2}{x+2}=\frac{(x+2)(x+1)}{x+2}=x+1$ Therefore, $\lim\limits_{x \to -2}\frac{x^{2}+3x+2}{x+2}=\lim\limits_{x \to -2}x+1=-2+1=-1$
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.