University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises - Page 111: 15

Answer

$$\lim_{x\to0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}=-\frac{1}{4}$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\frac{1}{2+x}-\frac{1}{2}}{x}=\lim_{x\to0}\frac{\frac{2-(2+x)}{2(2+x)}}{x}=\lim_{x\to0}\frac{\frac{-x}{2(2+x)}}{x}$$ $$A=\lim_{x\to0}\frac{-x}{2x(2+x)}=\lim_{x\to0}\frac{-1}{2(2+x)}$$ $$A=\frac{-1}{2(2+0)}=-\frac{1}{4}$$
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