University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 38

Answer

$\lim\limits_{x \to -1}\frac{\sqrt (x^{2}+8) - 3}{x+1} = -\frac{1}{3} = 0.33$

Work Step by Step

$\frac{\sqrt (x^{2}+8) - 3}{x+1}$ =$\frac{(\sqrt (x^{2}+8) - 3)(\sqrt (x^{2}+8) + 3)}{(x+1)(\sqrt (x^{2}+8) + 3)} $ = $\frac{(x^{2}+8)-9}{(x+1)(\sqrt (x^{2}+8) + 3)}$ = $\frac{(x^{2}-1)}{(x+1)(\sqrt (x^{2}+8) + 3)}$ = $\frac{(x-1)(x+1)}{(x+1)(\sqrt (x^{2}+8) + 3)} = \frac{(x-1)}{(\sqrt (x^{2}+8) + 3)} $ Now, $\lim\limits_{x \to -1}\frac{\sqrt (x^{2}+8) - 3}{x+1} =\lim\limits_{x \to -1} \frac{(x-1)}{(\sqrt (x^{2}+8) + 3)} = \frac{-1-1}{3+3} =-\frac{2}{6} = -\frac{1}{3} = -0.33$
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.