## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 16

#### Answer

$\dfrac{-1}{6}$

#### Work Step by Step

Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{\sin x-x}{x^3}$ Need to check that the limit has an indeterminate form. Thus, $f(0)=\dfrac{\sin (0)-0}{0^3}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to 0}\dfrac{\cos x -1}{3x^2}=\dfrac{0}{0}$ Again apply L-Hospital's rule: we get $\lim\limits_{x \to 0}\dfrac{-\sin x}{6x}=\dfrac{-\sin 0}{6(0)}=\dfrac{0}{0}$ Now, again apply L-Hospital's rule . we get $\lim\limits_{x \to 0}\dfrac{-\cos x}{6}=\dfrac{-\cos 0}{6}=\dfrac{-1}{6}$

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