Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 125: 24


We define $f(2)=3$

Work Step by Step

Consider the function $f(x)$ $f(x)=\frac{x^3-8}{x^2-4}$ $f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$ $f(x)=\frac{x^2+2x+4}{x+2}$ We see that $\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}\frac{x^2+2x+4}{x+2}=\frac{2^2+2\times2+4}{2+2}=3$ Since $f(x)$ is continuous at $2$ if and only if $\lim\limits_{x\to2}f(x)=f(2)$ Therefore, to make $f(x)$ continuous at $2$, we define $f(2)=3$.
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