Limits of integrals How would you show that if
$f : [0,1] \rightarrow \Bbb R$ is continuous, then
$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f\left( \frac{x_1+x_2+\cdots+x_n}{n} \right)~dx_1~dx_2\cdots dx_n = f\left( \frac{1}{2} \right) $$
and
$$\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f((x_1 x_2 \cdots x_n)^{\frac{1}{n}})~dx_1~dx_2\cdots dx_n = f\left(\frac{1}{e}\right) $$

The first result can be shown introducing a Dirac delta (and its integral representation) into the game. We have
$$
\lim_{n\rightarrow \infty}\int_0^1\int_0^1 \cdots \int_0^1 f \left( \frac{x_1+x_2+\cdots +x_n}{n} \right)~dx_1~dx_2\cdots dx_n =
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}} dt\ f(t)\int_0^1\int_0^1\cdots \int_0^1  \delta\left(t-\frac{1}{n}\sum_{k=1}^n x_k\right)~dx_1~dx_2\cdots dx_n=
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi} e^{i tz} \int_0^1\int_0^1 \cdots \int_0^1\ e^{-iz\sum_{k=1}^n x_k/n} ~dx_1~dx_2\cdots dx_n=
$$
$$
\lim_{n\rightarrow \infty}\int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\int_0^1 dx\ e^{-izx/n}\right]^n=\lim_{n\rightarrow \infty} \int_{\mathbb{R}}dt\ f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{i tz}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n
$$
and using
$$
\lim_{n\to\infty}\left[\frac{i n \left(-1+e^{-\frac{i z}{n}}\right)}{z}\right]^n=e^{-i z/2}
$$
the result is 
$$
\int_{\mathbb{R}}dt f(t)\int_{\mathbb{R}}\frac{dz}{2\pi}e^{iz(t-1/2)} =\int_{\mathbb{R}}dt\ f(t)\delta(t-1/2)=f(1/2)\ .
$$
