## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 23

#### Answer

We define $f(2)=3$

#### Work Step by Step

Consider the function $f(x)$ $f(x)=\frac{x^2-x-2}{x-2}$ $f(x)=\frac{(x-2)(x+1)}{x-2}$ $f(x)=x+1$ We see that $\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}(x+1)=2+1=3$ Since $f(x)$ is continuous at $2$ if and only if $\lim\limits_{x\to2}f(x)=f(2)=3$ Therefore, to make $f(x)$ continuous at $2$, we define $f(2)=3$.

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