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 1153: 29


$ \dfrac{1}{2}$

Work Step by Step

In order to to find the solution we will rationalize the function and then simplify. $\dfrac{\sqrt {1+x}-1}{x} \times \dfrac{\sqrt {1+x}+1}{\sqrt {1+x}+1}$ or, $=\dfrac{(1+x)-1}{x[\sqrt {1+x}+1]}$ or, $=\dfrac{1}{\sqrt {1+x}+1}$ Now, $\lim_\limits{x\to 0} \dfrac{\sqrt {1+x}-1}{x}=\lim_\limits{x\to 0} \dfrac{1}{\sqrt {1+x}+1}$ or, $= \dfrac{1}{\sqrt {1+0}+1}$ or, $= \dfrac{1}{2}$
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