calculus - What is the $lim_limits{nto infty}frac{sinfrac{1}{n}}{frac{1}{n}}$

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

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 $1$ ?

Answer

HINT:

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

So the limit becomes:

$$\lim_\limits{x\to 0^+}\frac{\sin x}{x}$$