Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 30

Answer

$\frac{-4}{5}$

Work Step by Step

$\lim\limits_{x \to -4}\frac{\sqrt {x^2+9}-5}{x+4}=\lim\limits_{x \to -4}\frac{\sqrt {x^2+9}-5}{x+4}*\frac{\sqrt {x^2+9}+5}{\sqrt {x^2+9}+5}=\lim\limits_{x \to -4}\frac{x^2+9-25}{(x+4)\sqrt {x^2+9}+5}=\lim\limits_{x \to -4}\frac{x^2-16}{(x+4)(\sqrt {x^2+9}+5)}=\lim\limits_{x \to -4}\frac{(x+4)(x-4)}{(x+4)(\sqrt {x^2+9}+5)}=\lim\limits_{x \to -4}\frac{x-4}{\sqrt {x^2+9}+5}=\frac{-4-4}{\sqrt {(-4)^2+9}+5}=\frac{-8}{\sqrt {16+9}+5}=\frac{-8}{5+5}=\frac{-4}{5}$
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