University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 108

Answer

The graph of the function is shown below.
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Work Step by Step

$$y=f(x)=\sin\Big(\frac{\pi}{x^2+1}\Big)$$ The graph is shown below. - First, we see from the formula that $f(x)$ is a sine function, whose range is $[-1,1]$, so we expect the graph to range only from $y=-1$ to $y=1$. We see from the graph that in fact the graph only ranges from $y=0$ to $y=1$. - To understand why the graph ranges only from $y=0$ to $y=1$, we look at the denominator $x^2+1$. We know that $$x^2+1\ge1$$ $$0\lt\frac{\pi}{x^2+1}\le\pi$$ $$0\lt\sin\Big(\frac{\pi}{x^2+1}\Big)\le1$$ (from $0$ to $\pi$, a sine function only ranges from $0$ to $1$). - Furthermore, we notice from the graph that as $x\to\pm\infty$, the graph approaches $0$. This is because as $x\to\pm\infty$, $x^2+1$ approaches $\infty$, so $\frac{\pi}{x^2+1}$ approaches $0$ and $\sin\Big(\frac{\pi}{x^2+1}\Big)$ approaches $0$ as well.
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