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

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?

• You know that $\sin t = t - \frac{t^3}{6} + O(t^5)$? – Daniel Fischer Oct 28 '14 at 23:14
• @DanielFischer yep, got it. thanks. – user159356 Oct 28 '14 at 23:16

As $$1 - \frac{\sin t}t \sim_{t\to 0} \frac{t^2}{3!} \\ \log (1+\epsilon) \sim_{\epsilon\to 0} \epsilon$$
you have $$\frac{\sin^{n}(\frac{x}{\sqrt{n}})}{\left(\frac{x}{\sqrt{n}} \right)^n} = \exp \left[ n\log \frac{\sin \frac x{\sqrt n}}{\frac x{\sqrt n}} \right] \to \exp \left[ n\left( -\frac 16 \left(\frac x{\sqrt n}\right)^2 \right) \right] = \exp\left( -\frac{x^2}6\right)$$
If you know that $$\sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,$$