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 55: 46


The limit does not exist

Work Step by Step

We have to estimate the limit: $\displaystyle\lim_{x\rightarrow 2\pm} \cos\left(\dfrac{\pi}{2}(x-[x])\right)$ Compute the value of $\cos\left(\dfrac{\pi}{2}(x-[x])\right)$ for values of $x$ approaching $2$ from left and right: $\cos\left(\dfrac{\pi}{2}(1.9-[1.9])\right)\approx 0.15643447$ $\cos\left(\dfrac{\pi}{2}(1.99-[1.99])\right)\approx 0.0.015707317$ $\cos\left(\dfrac{\pi}{2}(1.999-[1.999])\right)\approx 0.0015707957$ $\cos\left(\dfrac{\pi}{2}(1.9999-[1.9999])\right)\approx 0.0001570796$ $\cos\left(\dfrac{\pi}{2}(2.1-[2.1])\right)\approx 0.98768834$ $\cos\left(\dfrac{\pi}{2}(2.01-[2.01])\right)\approx 0.99987663$ $\cos\left(\dfrac{\pi}{2}(2.001-[2.001])\right)\approx 0.99999877$ $\cos\left(\dfrac{\pi}{2}(2.0001-[2.0001])\right)\approx 0.99999999$ Therefore we get: $\displaystyle\lim_{x\rightarrow 2^{-}} \cos\left(\dfrac{\pi}{2}(x-[x])\right)=0$ $\displaystyle\lim_{x\rightarrow 2^{+}} \cos\left(\dfrac{\pi}{2}(x-[x])\right)=1$ As the left hand limit and the right hand limit are not equal, the limit of the function in $x=2$ does not exist.
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