Problem with a sequence with multiple integrals How to compute the following limit, 
$\displaystyle \lim\limits_{n \to \infty} \int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n$ ?
I will greatly appreciate it if a solution only uses analysis. Thank you.
 A: This one is perhaps easier than the arbitrary function $f(x)$ in the example in the comments.
First note that:
$$
\begin{aligned}
&\int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n\\&=\Im{\left[\int_0^1 \int_0^1 \ldots \int_0^1 \text{exp} \bigg(i\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n\right]}\\
&=\Im{\left[\prod_{j=1}^{n}\int_0^1 \text{exp} \bigg(i\frac{x}{n}\bigg)\,dx \right]}\\
&=\Im{\left[ \prod _{j=1}^{n}in\left(1-\text{exp}\left[\frac {i}{n}\right]\right)
 \right]}\\
&=\Im{\left[ \left(in\left(1-\text{exp}\left[\frac {i}{n}\right]\right)\right)^n\
 \right]}\
\end{aligned}
$$
then take the limit:
$$
\begin{aligned}
\lim _{n\rightarrow \infty }\left(in\left(1-\text{exp}\left[\frac {i}{n}\right]\right)\right)^n&=\lim _{n\rightarrow \infty } \left( 2n\sin \left( \frac{1}{2n}
 \right)  \right) ^{n}\text{exp}\left[\frac{i}{2}\right]\\
&=\lim _{n\rightarrow \infty } \left(1-\frac{1}{24n^2}+\dots\right) ^{n}\text{exp}\left[\frac{i}{2}\right]\\
&=\text{exp}\left[\frac{i}{2}\right]
\end{aligned}$$
then take the imaginary part to get:

$$\displaystyle \lim\limits_{n \to \infty} \int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n=\sin{\frac{1}{2}}$$

which is the same answer you would get from the answer in the link in the comments.
