Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.6 Continuity - 2.6 Exercises - Page 108: 17

Answer

Not continuous

Work Step by Step

We are given the function: $f(x)=\begin{cases} \dfrac{x^2-1}{x-1},\text{ if }x\not=1\\ 3,\text{ if }x=3 \end{cases}$ We use the continuity checklist to determine if $f$ is continuous in $a=1$: 1) $f$ is defined for $a=1$. 2) $\lim\limits_{x \to 1^-} f(x)=\lim\limits_{x \to 1^-} \dfrac{x^2-1}{x-1}=\lim\limits_{x \to 1^-} \dfrac{(x-1)(x+1)}{x-1}$ $=\lim\limits_{x \to 1^-} (x+1)=1+1=2$ $\lim\limits_{x \to 1^+} f(x)=\lim\limits_{x \to 1^+} \dfrac{x^2-1}{x-1}=\lim\limits_{x \to 1^+} \dfrac{(x-1)(x+1)}{x-1}$ $=\lim\limits_{x \to 1^+} (x+1)=1+1=2$ $\lim\limits_{x \to 1^-} f(x)=\lim\limits_{x \to 1^+} f(x)$ Therefore $\lim\limits_{x \to 1} f(x)$ exists. 3) $f(1)=3$ $\lim\limits_{x \to 1}=2$ As $f(1)\not=\lim\limits_{x \to 1} f(x)$ condition 3 is not satisfied, the function is not continuous in $a=1$.
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