Find the limit of function I do not know how to approach this problem. Is there something related to integrating?

The function $f$ is defined on $[0,1]$, continuous, and positive. Find the following limit:
$$\lim_{n\to \infty} \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}$$

 A: Take log on both sides, 
Let $L_n= \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}\implies \ln L_n=\frac {\sum_{r=1}^n\ln f(\frac rn)}{n}$, which corresponds to Riemann sum for the integral $\int_0^1 \ln f(x) dx$. 
$\lim(\ln L_n)=\int_0^1 \ln f(x) dx\implies L_n\to \exp (\int_0^1 \ln f(x) dx)$ 
In this case, you may convert to Riemann sum like this also: Replace $1/n$ by $dx$, put $r/n =x$. When $r=1, x=1/n\to 0$ as $n\to \infty$ (This gives lower limit for the above integral) and when $r=n, x=n/n\to 1 $ as $n\to \infty$ (for the upper limit).
A: You have a product and powers. These are nasty; we like sums, sums are nice.
So take the logarithm to convert products into sums:
$$
\log \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}
= \frac{1}{n}\sum_{k=1}^n \log f\!\left(\frac{k}{n}\right) \tag{1}
$$
Now, this looks like a Riemann sum, applied to the continuous, well-defined (can you see why?) function $g = \log \circ f$. In particular,
$$
\lim_{n\to \infty}
\frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right) = 
\int_0^1 g(x)dx \tag{2}
$$
Now, since $\exp$ is continuous, we have
$$
\lim_{n\to \infty} e^{\frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right)}
= e^{\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right)}
= e^{\int_0^1 g(x)dx} \tag{3}
$$
i.e., your limit is
$$
\lim_{n\to \infty} \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)} 
= \boxed{e^{\int_0^1 \log(f(x))dx}}
$$

Sanity check: let's verify on a couple examples!

*

*$f$ is constant: if $f(x)=a>0$ for all $x$, then the limit is easily shown to be $\sqrt[n]{a^n}=a$, and $e^{\int_0^1 \log(a) dx} = e^{\log a} = a$. This checks out.


*$f(x)=x+1$ (this is continuous, positive, etc.). Then
$$
\sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)} 
= \sqrt[n]{ \frac{n+1}{n}\cdot\frac{n+2}{n}\cdots \frac{2n}{n} }
= \sqrt[n]{ \frac{(2n)!}{n!\cdot n^n} }
$$
and,, e.g., using Sterling's inequality, you can check the limit is $4e^{-1}$.
On the other hand,
$e^{\int_0^1 \log(x+1) dx} = e^{\log 4-1} = 4e^{-1}$. This  again checks out.
