I am confused by the following limit:
When I graph it on my calculator I can see the limit goes towards 0, but how do I mathematically do this problem?
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$\begingroup$ $-1\leq \sin(5x) \leq 1$. $\endgroup$– amWhyFeb 23, 2020 at 17:22
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$\begingroup$ Would be more interesting to take the limit of this ratio as $x\rightarrow 0.$ $\endgroup$– mjwFeb 23, 2020 at 18:09
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3 Answers
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This is my first time posting here. Please excuse the formatting!
Since this question comes from a calculus course, you might want to learn this a little more intuitively. We know that $-1\leq \sin(5x) \leq 1$ (since changing the period doesn't affect the amplitude). Since we're considering $x\to\infty$ ("eventually" positive), we don't have to worry about flipping around the inequalities when we multiply everything by $1/x$. The goal now is to get the inner term into the limit form you posted:
$$-1\leq \sin(5x) \leq 1$$ $$\frac{-1}{x}\leq \frac{\sin(5x)}{x} \leq\frac{1}{x}$$ $$\lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(5x)}{x}\leq\lim_{x\to\infty}\frac{1}{x}$$ $$0\leq\lim_{x\to\infty}\frac{\sin(5x)}{x}\leq0$$
Looking at the last inequality, the only possibility that remains is that $$\lim_{x\to\infty}\frac{\sin(5x)}{x}=0.$$
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A less rigorous but more intuitive explanation is that $\sin(5x)$ is bounded by $-1$ and $1$, so it won't matter for extremely large values of $x$. The only thing that changes significantly is the denominator, so you're comparing some value oscillating between $-1$ and $1$ to some extremely large number. Therefore, the limit is $0$.
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Note that,
$$\lim_{x\to\infty} \frac{\sin 5x}x \le \lim_{x\to\infty} \frac{1}x=0$$
Similarly,
$$\lim_{x\to\infty} \frac{\sin 5x}x \ge -\lim_{x\to\infty} \frac{1}x=0$$
Thus,
$$\lim_{x\to\infty} \frac{\sin 5x}x=0$$
