I would like to calculate the following limit: $\lim_ {n \to \infty} {\left( {n\cdot \sin{\frac{1}{n}}} \right)^{n^2}}$ I would like to calculate the following limit: $$\lim_ {n \to \infty} {\left( {n\cdot \sin{\frac{1}{n}}} \right)^{n^2}}$$
or prove that it does not exist. Now I know the result is $\frac{1}{\sqrt[6]{e}}$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
 A: Detailed derivation, using Taylor expansions (you will need $\sin x = x-\frac{x^3}{6} + o(x^3)$ and $\ln(1+x) = x + o(x)$, for $x\to 0$):


*

*Rewrite the quantity in the (more convenient) exponential form:
$$
\left(n\sin \frac{1}{n}\right)^{n^2} = e^{n^2 \ln\left(n\sin \frac{1}{n}\right)}
$$

*Use the Taylor expansion of $\sin$ around $0$ (as $\frac{1}{n}\to 0)$:

 $$\left(n\sin \frac{1}{n}\right)^{n^2} = e^{n^2 \ln\left(n\left(\frac{1}{n}-\frac{1}{6n^3}+o\left(\frac{1}{n^3}\right)\right)\right)}= e^{n^2 \ln\left(1-\frac{1}{6n^2}+o\left(\frac{1}{n^2}\right)\right)}$$

(I am hiding the computations to give only the hint, but place your mouse over the gray area to reveal them)


*

*Use the Taylor expansion of $\ln(1+x)$ around $0$ on the result (as now $\frac{1}{6n^2}\to 0)$:

$$\left(n\sin \frac{1}{n}\right)^{n^2} = e^{n^2 \left(-\frac{1}{6n^2}+o\left(\frac{1}{n^2}\right)\right)}= e^{-\frac{1}{6}+o\left(1\right)} $$


*This gives you the limit:
$$
e^{-\frac{1}{6}+o\left(1\right)} \xrightarrow[n\to\infty]{} e^{-\frac{1}{6}}
$$
where all $o(\cdot)$'s are taken with regard to $n$ going to $\infty$.
A: $$
\left( {n\cdot \sin{\frac{1}{n}}} \right)^{n^2}
= e^{n^2 \log (n\sin(1/n))}
$$
now
$$
\log (\sin(x)/x) = \log\frac{x - x^3/6 + o(x^3)}{x}
= \log(1-x^2/6+o(x^2)) = -x^2/6 + o(x^2)
$$
hence
$$
e^{n^2 \log (n\sin(1/n))} = e^{n^2 (-\frac{1}{6n^2}+o(1/n^2))} = e^{-1/6 + o(1)} \to e^{-1/6}
$$
