Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 78: 82



Work Step by Step

We have to find function $f(x)$ so that $\lim\limits_{x \to 1}\left(\dfrac{f(x)}{x-1}\right)=2$. In order for the limit to exist $f(x)$ should have a factor $(x-1)$. We can choose any function $g$ so that we have: $f(x)=(x-1)g(x)$ $g(1)=2$ For example: $g(x)=(x+1)\Rightarrow f(x)=(x-1)(x+1)=x^2-1$ $\lim\limits_{x \to 1}\left(\dfrac{x^2-1)}{x-1}\right)=\lim\limits_{x \to 1}\left(\dfrac{(x-1)(x+1)}{x-1}\right)=\lim\limits_{x \to 1} (x+1)=1+1=2$
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