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$$\lim_\limits{n\to \infty}\frac{\sin\frac{1}{n}}{\frac{1}{n}}$$

How am I supposed to know this equals 1?

I could sub $x= \frac{1}{n}$ to get

$$\lim_\limits{x\to \infty}\frac{\sin(x)}{x} $$

Using L'Hopital's I'd get:

$$\lim_\limits{x\to \infty}\frac{\cos(x)}{1} $$

But, $\cos(x)$ just cycles between $-1$ and $1$, so how can the limit be $1$ ?

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    $\begingroup$ You should replace $\infty$ with $0$. $\endgroup$ – S.C.B. Mar 22 '16 at 13:54
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    $\begingroup$ Your second displayed line is wrong. If $x=\frac1n$, then $x\to 0$ as $n\to\infty$. $\endgroup$ – Brian M. Scott Mar 22 '16 at 13:54
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    $\begingroup$ For the record: L'Hopital here is overkill (and circular). You use the fact that $\sin^\prime=\cos$ to apply L'Hopital, but then you have by definition $\sin^\prime(0) = \lim_{x\to 0} \frac{\sin x - \sin 0}{x-0} = \frac{\sin x}{x}$ already. So applying L'Hopital's rule here is tantamount to opening a can with a jackhammer. $\endgroup$ – Clement C. Mar 22 '16 at 14:29
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HINT:

If $x=\frac{1}{n}$, then as $n\to\infty$, $x\to 0^+$

So the limit becomes: $$\lim_\limits{x\to 0^+}\frac{\sin x}{x}$$

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$$\lim_{n\to \infty}\frac{\sin(1/n)}{1/n}=\lim_{x \to 0^+}\frac{\sin(x)}{x}=1$$

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