Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 36

Answer

$\lim\limits_{x\to\pi}\sin(x+\sin x)=0$

Work Step by Step

$$A=\lim\limits_{x\to\pi}\sin(x+\sin x)$$ Let $f(x)=\sin x$ and $g(x)=x+\sin x$, we have $$A=\lim\limits_{x\to\pi}f(g(x))$$ Since $\sin x$ is continuous on $R$, we can apply Theorem 8: $$A=f(\lim\limits_{x\to\pi}g(x))$$ $$A=\sin(\lim\limits_{x\to\pi}(x+\sin x))$$ $$A=\sin(\pi+\sin\pi)$$ $$A=\sin(\pi+0)$$ $$A=\sin\pi$$ $$A=0$$
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