## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

# Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 925: 14

#### Answer

The function $f(x)$ is not continuous at $x=-2$.

#### Work Step by Step

$f(x)=\dfrac{x^2-4}{x+2}$ We need to check whether the function $f(x)$ is continuous at $x=-2$. We check the left-hand and right-hand limits. If they are equal to each other and the value of the function at the limit, then the function is continuous at that point. Since, $f(-2)=4$ $\lim\limits_{x \to -2}f(x)=\lim\limits_{x \to -2}\dfrac{x^2-4}{x+2} \\=\lim\limits_{x \to -2}\dfrac{(x+2)(x-2)}{x+2} \\=\lim\limits_{x \to -2}(x-2)\\=(-2)-2\\=-4$ We can see that the function $f(x)$ at $x=-2$ is $4 \ne -4$, so the function cannot be defined at $x=-2$. Therefore, the function $f(x)$ is not continuous at $x=-2$.

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