## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 39

#### Answer

1) Check the continuity of $f(x)=1-x^2$ 2) Check the continuity of $f(x)=\ln x$ 3) Check the continuity of $f(x)$ at $1$.

#### Work Step by Step

$1-x^2$ is continuous on $R$, so it is continuous on $(-\infty,1)$ $\ln x$ is continuous on $(0,\infty)$, so it is continuous on $(1,\infty)$ Therefore, $f(x)$ is continuous on $(-\infty,1)\cup(1,\infty)$ Now we only need to check if $f(x)$ is continuous at $1$. By definition, $f(x)$ is continuous at $1$ if and only if $$\lim\limits_{x\to1}f(x)=f(1)$$ 1) For $x\lt;1$, we have $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(1-x^2)=1-1^2=0$ 2) For $x\gt1$, we have $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\ln x=\ln1=0$ Since $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)=0$ Therefore, $\lim\limits_{x\to1}f(x)=0$ Notice that $f(1)=1-1^2=0$ Since $\lim\limits_{x\to1}f(x)=f(1)$, $f(x)$ is continuous at $1$. Overall, $f(x)$ is continuous on $(-\infty,\infty)$.

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