Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 102: 2


a) $\lim\limits_{x\to 2}[f(x)+g(x)]=1$ b) $\lim\limits_{x\to 0}[f(x)-g(x)]$ does not exist. c) $\lim\limits_{x\to -1}[f(x)g(x)]=2$ d) $\lim\limits_{x\to 3}\frac{f(x)}{g(x)}$ does not exist. e) $\lim\limits_{x\to 2}[x^2f(x)]=-4$ f) $f(-1)+\lim\limits_{x\to -1}g(x)=5$

Work Step by Step

a) $\lim\limits_{x\to 2}f(x)=-1$ and $\lim\limits_{x\to 2}g(x)=2$ So, $\lim\limits_{x\to 2}[f(x)+g(x)]=\lim\limits_{x\to 2}f(x)+\lim\limits_{x\to 2}g(x)=-1+2=1$ b) $\lim\limits_{x\to 0}f(x)=2$ But $\lim\limits_{x\to 0}g(x)$ does not exist since $\lim\limits_{x\to 0^+}g(x)=1$ while $\lim\limits_{x\to 0^-}g(x)=3$ Therefore, we calculate the one-side limit: $\lim\limits_{x\to 0^+}[f(x)-g(x)]=2-1=1$ $\lim\limits_{x\to 0^-}[f(x)-g(x)]=2-3=-1$ So, $\lim\limits_{x\to 0}[f(x)-g(x)]$ does not exist. c) $\lim\limits_{x\to -1}f(x)=1$ and $\lim\limits_{x\to -1}g(x)=2$ So, $\lim\limits_{x\to -1}[f(x)g(x)]=\lim\limits_{x\to -1}f(x)\times\lim\limits_{x\to -1}g(x)=1\times2=2$ d) $\lim\limits_{x\to 3}f(x)=1$ and $\lim\limits_{x\to 3}g(x)=0$ $\lim\limits_{x\to 3}[\frac{f(x)}{g(x)}]=\frac{\lim\limits_{x\to 3}f(x)}{\lim\limits_{x\to 3}g(x)}$ However, $\lim\limits_{x\to 3}g(x)=0$ Therefore, $\lim\limits_{x\to 3}\frac{f(x)}{g(x)}$ is undefined and does not exist. e) $\lim\limits_{x\to 2}f(x)=-1$ and $\lim\limits_{x\to 2}x^2=2^2=4$ $\lim\limits_{x\to 2}[x^2f(x)]=\lim\limits_{x\to 2}x^2\times\lim\limits_{x\to 2}f(x)=(-1)\times4=-4$ f) $f(-1)=3$ and $\lim\limits_{x\to -1}g(x)=2$ $f(-1)+\lim\limits_{x\to -1}g(x)=3+2=5$
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