Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

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

Answer

$-\dfrac{1}{3}$

Work Step by Step

We want to find $\lim\limits_{x \to -1}\dfrac{\sqrt{x^2+8}-3}{x+1}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -1 is zero, and if we try to substitute $x=-1$ directly, we get zero in both the numerator and the denominator. But note that $\dfrac{\sqrt{x^2+8}-3}{x+1}=\left(\dfrac{\sqrt{x^2+8}-3}{x+1}\right)\left(\dfrac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\right)=\dfrac{x^2+2-9}{(x+1)(\sqrt{x^2+8}+3)}=\dfrac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\dfrac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\dfrac{x-1}{\sqrt{x^2+8}+3}$ for all $x \neq -1.$ Thus $\lim\limits_{x \to -1}\dfrac{\sqrt{x^2+8}-3}{x+1}=\lim\limits_{x \to -1}\dfrac{x-1}{\sqrt{x^2+8}+3}=\dfrac{-1-1}{\sqrt{1+8}+3}=\dfrac{-2}{\sqrt{9}+3}=\dfrac{-2}{3+3}=\dfrac{-2}{6}=-\dfrac{1}{3}.$
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