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!
 A: hint: $n \sin\left(\dfrac{x}{n}\right) = x\cdot \dfrac{\sin\left(\dfrac{x}{n}\right)}{\dfrac{x}{n}}$
A: We have
$$\mathop {\lim }\limits_{n \to \infty } \frac{n}{{{e^x} - 1}}\sin \left( {\frac{x}
{n}} \right) = \frac{1}{{{e^x} - 1}}\mathop {\lim }\limits_{n \to \infty } n\sin \left( {\frac{x}{n}} \right) = \frac{x}{{{e^x} - 1}}\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{x}{n}} \right)}}{{\frac{x}{n}}} = \frac{x}
{{{e^x} - 1}}$$
since
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{x}
{n}} \right)}}
{{\frac{x}
{n}}} = \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}
{u} = 1\,\,\,\left( {u = \frac{x}
{n}} \right)$$
A: Yet another way.
Let $x \neq 0$. Note that $\sin \frac{x}{n} = \frac{x}{n} - \frac{(x/n)^{3}}{3!} + o(1)$ as $n \to \infty$. But then
$$
\frac{n\sin \frac{x}{n}}{e^{x} - 1} = \frac{x - \frac{x^{3}}{3!n^{2}} + o(1)}{e^{x}-1} \to \frac{x}{e^{x}-1}
$$
as $n \to \infty$.
A: With equivalents, this is trivial: $\;\sin \dfrac xn\sim \dfrac xn\enspace(n\to\infty)$, hence
$$\frac n{\mathrm e^x -1}\,\sin \frac xn\sim\frac n{\mathrm e^x -1}\frac xn=\frac x{\mathrm e^x -1}.$$
