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

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 910: 59

Answer

$f(x)$ is continuous at $0$.

Work Step by Step

When $f(0)=\lim\limits_{x\to 0} f(x)$, then $f(x)$ will be continuous at $x=0$. We have: $\lim\limits_{x\to 0^{-}}f(x)=\lim\limits_{x\to 0^{-}} 2e^x =2e^{0}=2*1=2$ and $\lim\limits_{x\to 0^{+}}f(x)=\lim\limits_{x\to 0^{+}}\dfrac{x^2 (x+2)}{x^2} =2+0=2$ Thus: $\lim\limits_{x\to 0}f(x)=2$ Since, $f(0)=2$, we can see that our result satisfies the statement because $2 = 2$. Therefore, $f(x)$ is continuous at $0$.
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