Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2: 47

Answer

$\dfrac{1}{3}$

Work Step by Step

Note that, $\lim\limits_{x \to a}\sin x=\sin a$ and $\lim\limits_{x \to a}\cos x=\cos a$ for all real numbers $a$. Using this along with our limit laws, we get $\lim\limits_{x \to 0}\dfrac{1+x+\sin x}{3\cos x}=\dfrac{1+0+\sin 0}{3\cos 0}=\dfrac{1+0+0}{3(1)}=\dfrac{1}{3}.$
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