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Show that

$$\lim_{x\to \infty} \sin\left(\frac 1x\right) = 0$$

I don't even really know where to start on this question. I know that the limit definition at infinity is: for all $\epsilon>0$, there exists $R>0$ such that $x>R$ implies that $\left|f(x) - L\right| < \epsilon$.

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If I make $x$ bigger and bigger, what happens to $\frac 1 x$? I assume it's clear to you that it goes towards zero.

Now if I make the $x$ in $\sin x$ go towards zero, what happens to $\sin x$?

What I'm getting at is that

$$\lim_{x\to \infty} \sin\left(\frac 1 x\right)=\sin\left(\lim_{x\to\infty}\frac 1 x\right)$$

Which should be obvious if you have a good understanding of the concept of a limit.

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First, as $x \to \infty$, $\frac1{x} \to 0$.

Second, for $|x| < \pi/2$, $|\sin(x)| \le |x|$.

Therefore, for $x > \frac{2}{\pi}$, $|\sin(\frac1{x})| \le \frac1{x} $, so $\sin(\frac1{x}) \to 0$.

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