Limit $n \rightarrow \infty \frac{n}{e^x-1} \sin\frac{x}{n}$ I am just working through some practice questions and cannot seem to get this one. Plugging this into wolfram alpha I know the limit should be $\frac{x}{e^x-1}$, but I am having a bit of trouble working through this.
I have tried to rewrite the limit as $\frac{\frac{n}{e^x-1}}{\csc(\frac{x}{n})}$, then applying L'Hopital's rule a few times gets pretty messy.
Is there an easier what to compute this limit that I am not seeing?
Thanks very much in advance!

With equivalents, this is trivial: $\;\sin \dfrac xn\sim \dfrac xn\enspace(n\to\infty)$, hence
$$\frac n{\mathrm e^x -1}\,\sin \frac xn\sim\frac n{\mathrm e^x -1}\frac xn=\frac x{\mathrm e^x -1}.$$
