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I am wondering about a limit that wolframalpha got me and that you can find here wolframalpha

It says that $$\lim_{n \rightarrow \infty} \frac{\sin^{n}(\frac{x}{\sqrt{n}})}{\left(\frac{x}{\sqrt{n}} \right)^n} = e^{-\frac{x^2}{6}} $$

Does anybody know if there is a "easy" way to get this?

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    $\begingroup$ You know that $\sin t = t - \frac{t^3}{6} + O(t^5)$? $\endgroup$ – Daniel Fischer Oct 28 '14 at 23:14
  • $\begingroup$ @DanielFischer yep, got it. thanks. $\endgroup$ – user159356 Oct 28 '14 at 23:16
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As $$ 1 - \frac{\sin t}t \sim_{t\to 0} \frac{t^2}{3!} \\ \log (1+\epsilon) \sim_{\epsilon\to 0} \epsilon $$

you have $$ \frac{\sin^{n}(\frac{x}{\sqrt{n}})}{\left(\frac{x}{\sqrt{n}} \right)^n} = \exp \left[ n\log \frac{\sin \frac x{\sqrt n}}{\frac x{\sqrt n}} \right] \to \exp \left[ n\left( -\frac 16 \left(\frac x{\sqrt n}\right)^2 \right) \right] = \exp\left( -\frac{x^2}6\right) $$

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Hints:

If you know that $$\sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,$$

then it will be helpful for you.

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