Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 39


$\displaystyle\lim_{x\rightarrow 0^+} \dfrac{\sin x}{|x|}=-1$ $\displaystyle\lim_{x\rightarrow 0^-} \dfrac{\sin x}{|x|}=1$ The limit in 0 does not exist

Work Step by Step

We have to estimate the limit: $\displaystyle\lim_{x\rightarrow \pm0} \dfrac{\sin x}{|x|}$ Compute $\dfrac{\sin x}{|x|}$ for values of $x$ approaching 0 from the left and from the right side: $\dfrac{\sin (-0.1)}{|-0.1|}\approx -0.99833417$ $\dfrac{\sin (-0.01)}{|-0.01|}\approx -0.99998333$ $\dfrac{\sin (-0.001)}{|-0.001|}\approx -0.99999983$ $\dfrac{\sin (0.1)}{|0.1|}\approx 0.99833417$ $\dfrac{\sin (0.01)}{|0.01|}\approx 0.99998333$ $\dfrac{\sin (0.001)}{|0.001|}\approx 0.99999983$ Therefore we got: $\displaystyle\lim_{x\rightarrow 0^+} \dfrac{\sin x}{|x|}=-1$ $\displaystyle\lim_{x\rightarrow 0^-} \dfrac{\sin x}{|x|}=1$ As the left hand limit and the right hand limit are not equal, the limit of the function in 0 does not exist.
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.