Answer
$\lim\limits_{x\to0}\dfrac{1-\cos{x}}{\sin{x}}=0.$
Work Step by Step
Using Theorem $1.9:$
$\lim\limits_{x\to0}\dfrac{1-\cos{x}}{\sin{x}}=\lim\limits_{x\to0}\dfrac{1-\cos{x}}{x}\times\lim\limits_{x\to0}\dfrac{x}{\sin{x}}=0\times1=0.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 1 - Limits and Their Properties - Review Exercises - Page 91: 25
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