Finding the limit of the following expression Given that $f$ is continuous and non-negative on $[0,1]$, I was asked to find the limit of:
$$\displaystyle\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^{n} dx\right)^{\frac{1}{n}}$$
 A: Let $M=\sup_{x\in[0,1]}f(x)$. It's not difficult to see that 
$$M\geq \displaystyle\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^{n} dx\right)^{\frac{1}{n}}.$$
On the other hand, there exists $x_0\in[0,1]$ such that $f(x_0)=M$ since $f$ is continuous on $[0,1]$. Again, by the continuity of $f$, for any $\epsilon>0$, there exists $\delta>0$ such that 
$$\tag{1}f(x)>M-\epsilon\mbox{ whenever }x\in (x_0-\delta,x_0+\delta)\cap[0,1].$$
Hence, we have
$$\tag{2}\int_0^1(f(x))^{n} dx\geq \int_{(x_0-\delta,x_0+\delta)\cap[0,1]}(f(x))^{n} dx\geq m((x_0-\delta,x_0+\delta)\cap[0,1])(M-\epsilon)^n,$$
where the first inequality follows from the assumption that $f$ is nonnegative, and the second inequality follows from $(1)$. Also, $m((x_0-\delta,x_0+\delta)\cap[0,1])$ is the "length" of the interval $(x_0-\delta,x_0+\delta)\cap[0,1]$. For example, if $(x_0-\delta,x_0+\delta)\subset[0,1]$, then $m((x_0-\delta,x_0+\delta)\cap[0,1])=2\delta$; if $x_0=0$, then $m((x_0-\delta,x_0+\delta)\cap[0,1])=\delta$. In any case, we always have 
$$m((x_0-\delta,x_0+\delta)\cap[0,1])\geq\delta>0.$$
(This is important when we take $n$-root and as $n$ goes to infinity.)
Now taking $n$-th root of $(2)$, you will get the answer. I think you can work out the details from here. 
