Prove $\lim_{n \to \infty}\int_0^1 \dots \int_0^1 f(\sqrt[n]{x_1\dots x_n})dx_1\dots dx_n = f(\frac{1}{e}).$ $f$ is continuous on $[0;1].$ $f$ is continuous on $[0;1].$ Prove $$\lim_{n \to \infty} \underbrace{\int_0^1 \cdots \int_0^1}_{n} f(\sqrt[n]{x_1\cdots x_n})\mathrm \, dx_1\cdots \mathrm dx_n = f(\frac{1}{e}).$$
At first, I thought that we should get inside the function with the limit, but it's probably restricted due to $dx_1\dots dx_n$. I feel like I'm missing an important Theorem here. And yet it seems like the problem should be easy. Can somebody smart help me out here (at least with a hint) ?
Perhaps, it's an induction method problem.
 A: A short solution could be. Consider the random variables $u_1,...,u_n$  which they are i.i.d with density function $\rho(u)=e^{-u}$ (exponential distribution with $\lambda =1 $). So the expected value $E[u] = \int^{\infty}_{0}u\cdot e^{-u}du = 1$. By law of large numbers $\frac{S_n}{n}\to E[u]=1$. Where $S_n := u_1+..+u_n$. Observe that 
$$\int_{0}^1\dots\int_{0}^1f(\sqrt[n]{x_1...x_n})dx_1\dots dx_n = \int_{0}^{\infty}\dots\int_{0}^{\infty}f(e^{-\frac{S_n}{n}}) e^{-u_1}du_1...e^{-u_n}du_n =E[f(e^{-\frac{S_n}{n}})]$$
Finally by continuity and dominated convergence
$$E[f(e^{-\frac{S_n}{n}})] \xrightarrow{n} E[f(e^{-1})]=f(e^{-1}).$$ 
A: Here's an elementary proof. Suppose $f(x) = x^k$ then we see that
\begin{align}
\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n})=&\ \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \left(\sqrt[n]{x_1\cdots x_n} \right)^k\\
=&\ \left(\int^1_0 dx\ x^{\frac{k}{n}} \right)^n = \left(1+\frac{k}{n} \right)^{-n}.
\end{align}
In particular, it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = \lim_{n\rightarrow \infty}\left(1+\frac{k}{n} \right)^{-n} = \left(\frac{1}{e}\right)^k.
\end{align}
If $f$ is a polynomial, then it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = f\left( \frac{1}{e}\right).
\end{align}
Next, if $f$ is continuous, then, by Wierestrass approximation theorem, there exists a sequence of polynomials $p_m$ such that $p_m \rightarrow f$ uniformly on $[0, 1]$. Finally, it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) =&\ \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \lim_{m\rightarrow \infty}p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\  \lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\ \lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\ \lim_{m\rightarrow \infty} p_m\left(\frac{1}{e} \right) = f\left(\frac{1}{e} \right).
\end{align}
