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: 31


$\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}=0$

Work Step by Step

We have to estimate the limit: $\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}$ Compute $\sin h\cos\dfrac{1}{h}$ for values of $h$ close to 0: $\sin (-0.01)\cos\dfrac{1}{-0.01}\approx -0.00862304$ $\sin (-0.001)\cos\dfrac{1}{-0.001}\approx -0.00056238$ $\sin (-0.0001)\cos\dfrac{1}{-0.0001}\approx 0.00009522$ $\sin (-0.00001)\cos\dfrac{1}{-0.00001}\approx 0.00000999$ $\sin (0.00001)\cos\dfrac{1}{0.00001}\approx -0.00000999$ $\sin (0.0001)\cos\dfrac{1}{0.0001}\approx -0.00009522$ $\sin (0.001)\cos\dfrac{1}{0.001}\approx 0.00056238$ $\sin (0.01)\cos\dfrac{1}{0.01}\approx 0.00862304$ Therefore we got: $\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}=0$
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