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: 1


a) $\lim\limits_{x \to 2}[f(x)+5g(x)]=-6$ b) $\lim\limits_{x \to 2}[g(x)]^3=-8$ c) $\lim\limits_{x \to 2}\sqrt{f(x)}=2$ d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}=-6$ e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}$ does not exist f) $\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}=0$

Work Step by Step

a) $\lim\limits_{x \to 2}[f(x) + 5g(x)]=\lim\limits_{x \to 2}f(x)+5\lim\limits_{x \to 2}g(x)=4+5\times(-2)=-6$ b) $\lim\limits_{x \to 2}[g(x)]^3=[\lim\limits_{x \to 2}g(x)]^3=(-2)^3=-8$ c) $\lim\limits_{x \to 2}\sqrt {f(x)}=\sqrt{\lim\limits_{x \to 2}f(x)}=\sqrt4=2$ d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}=\frac{3\times\lim\limits_{x \to 2}f(x)}{\lim\limits_{x \to 2}g(x)}=\frac{3\times4}{-2}=-6$ e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}=\frac{\lim\limits_{x \to 2}g(x)}{\lim\limits_{x \to 2}h(x)}$ However, we notice that $\lim\limits_{x \to 2}h(x)=0$ Therefore, this limit is undefined and as a result, does not exist. f) $\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}=\frac{\lim\limits_{x \to 2}g(x)\times\lim\limits_{x \to 2}h(x)}{\lim\limits_{x \to 2}f(x)}=\frac{(-2)\times 0}{4}=0$
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