Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1154: 54



Work Step by Step

$(f o g) (x)=\dfrac{4}{1/x+2-1}=\dfrac{4(x+2)}{-1-x}$ Now, $\lim_\limits{x \to 1} (f o g) (x)=\dfrac{4(1+2)}{-1-1}=-6$ and $( g o f) (x)=\dfrac{1}{\dfrac{4}{x-1}-1}=\dfrac{x-1}{4+2(x-1)}$ Now, $\lim_\limits{x \to 1} ( g o f) (x)=\dfrac{1-1}{4+2(1-1)}=0$ Answers: $-6,0$
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