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

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 $1$ ?

  • 1
    $\begingroup$ You should replace $\infty$ with $0$. $\endgroup$ – S.C.B. Mar 22 '16 at 13:54
  • 4
    $\begingroup$ 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
  • 2
    $\begingroup$ 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


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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.