Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises: 32

Answer

$\lim\limits_{x \to \infty} \frac{\sin x}{x-\pi} = 0$

Work Step by Step

As $\lim\limits_{x \to \infty} \frac{sinx}{x-\pi} $ the term $\sin x$ oscillates between 1 and -1. So the limit can be simplified into two different terms when infinity is substituted for x $-\frac{1}{\infty}$, and $\frac{1}{\infty}$, Both of which go to zero. So, it can be concluded that $\lim\limits_{x \to \infty} \frac{\sin x}{x-\pi} = 0$
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