University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 153: 68

Answer

See below for detailed explanations.
1538410916

Work Step by Step

The graphs are enclosed below. As seen from the graph, the graph of $y=(\sin x)/x$ appears to cross the $y-$axis at $y=1$, the graph of $y=(\sin 2x)/x$ appears to cross the $y-$axis at $y=2$, and the graph of $y=(\sin 4x)/x$ appears to cross the $y-$axis at $y=4$. But the table on the left side shows that for each case, as $x=0$, the corresponding value of $y$ is always undefined, which means the graphs do not really intersect the axis. Taken from the pattern viewed here, we also expect the graphs of $y=(\sin5x)/x$ to approach $5$ and $y=(\sin-3x)/x$ to approach $-3$ as $x\to0$ too. In fact, the whole pattern can be generalized to the case of $y=(\sin kx)/x$. In short, $$\lim_{x\to0}\frac{\sin kx}{x}=k$$ The reason for this can be traced to the fact that $\lim_{x\to0}\frac{\sin x}{x}=1$. $$\lim_{x\to0}\frac{\sin kx}{x}=k\lim_{x\to0}\frac{\sin kx}{kx}$$ If we take $kx=n$, then as $x\to0$, $n=kx$ also approaches $0$, meaning that $$\lim_{x\to0}\frac{\sin kx}{x}=k\lim_{n\to0}\frac{\sin n}{n}=k\times1=k$$
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