University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 153: 66



Work Step by Step

$$\lim_{h\to0}\frac{|0+h|-|0-h|}{2h}$$ $$=\lim_{h\to0}\frac{|h|-|-h|}{2h}$$ - As $h\to0^+$, meaning $h\gt0$ and $-h\lt0$, we have $|h|=h$ and $|-h|=-(-h)=h$ So, $$\lim_{h\to0^+}\frac{|h|-|-h|}{2h}=\lim_{h\to0^+}\frac{h-h}{2h}=\lim_{h\to0^+}\frac{0}{2h}=\lim_{h\to0^+}0=0$$ - As $h\to0^-$, meaning $h\lt0$ and $-h\gt0$, we have $|h|=-h$ and $|-h|=-h$ So, $$\lim_{h\to0^-}\frac{|h|-|-h|}{2h}=\lim_{h\to0^-}\frac{-h-(-h)}{2h}=\lim_{h\to0^-}\frac{0}{2h}=\lim_{h\to0^-}0=0$$ The left-side and right-side limits are equal with each other, therefore, $$\lim_{h\to0}\frac{|0+h|-|0-h|}{2h}=0$$ As mentioned, this limit really exists while $f(x)=|x|$ has no derivative at $x=0$.
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