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: 30


$ \dfrac{1}{8}$

Work Step by Step

In order to to find the solution we will rationalize the function and then simplify. $\dfrac{\sqrt {16+x}-4}{x} \times \dfrac{\sqrt {16+x}+4}{\sqrt {16+x}+4}$ or, $=\dfrac{(16+x)-16}{x[\sqrt {16+x}+4]}$ or, $=\dfrac{1}{\sqrt {16+x}+4}$ Now, $\lim_\limits{x\to 0} \dfrac{\sqrt {16+x}-4}{x}=\lim_\limits{x\to 0} \dfrac{1}{\sqrt {16+x}+16}$ or, $= \dfrac{1}{\sqrt {16}+4}$ or, $= \dfrac{1}{8}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.