Prove the following equation(much needed help): Let there be a given function $f \in C([0,1])$, $f(x)>0$; $x\in [0,1]$. Prove 
$$\lim_{n\to\infty} \sqrt[n]{f\left({1\over n}\right)f\left({2\over n}\right)\cdots f\left({n\over n}\right)}=e^{\int_0^1 \log f(x) \, dx} $$
All the questions before this required solving an definite integral without Newton Leibnitz formula, then this came up, can anyone provide help?
 A: $f$ is positive, and the logarithm is continuous on $(0,\infty)$, so we can take logarithms of both sides and swap the limit and logarithm to find
$$ \log{\left( \lim_{n\to\infty} \sqrt[n]{ f(1/n) f(2/n) \dotsm f(n/n) }\right)} = \lim_{n\to\infty} \log{\sqrt[n]{f(1/n)f(2/n)\dotsm f(n/n)}} $$
Now applying properties of the logarithm, the expression inside the limit on the right hand side is equal to
$$ \frac{1}{n}\log{\left( f(1/n)f(2/n)\dotsm f(n/n) \right)} = \frac{1}{n} \left( \log{f(1/n)}+\log{f(2/n)}+\dotsb+\log{f(n/n)} \right) \\
= \sum_{k=1}^{n} \frac{1}{n}\log{f(k/n)}, $$
which is a Riemann sum for $\log{f}$ on the interval $[0,1]$. Because $f$ is continuous, it is Riemann-integrable, and hence this sum must converge to the integral
$$ \int_0^1 \log{f(x)} \, dx, $$
which is the logarithm of the right-hand side of your original expression.
A: Your limit is the same as
$$\exp\{\lim_{n\to \infty} \frac{1}{n}\left(\log f(1/n) +\log f(2/n) + \cdots + \log f(n/n)\right)\}$$
and the limit inside converges to $\int_0^1 \log f(x)\, dx$, since the sum inside is a sequence of Riemann sums of the continuous function $\log f(x)$ over $[0,1]$.
