Finding $\lim\limits_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$ for continuous $f:[0,1]\to[0,\infty)$ 
Find $$\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$$if $f:[0,1]\rightarrow(0,\infty)$ is a continuous function.

My attempt:
Say $f(x)$ has a max. value $M$. Then $$\left(\int_0^1(f(x))^ndx\right)^\frac{1}{n}\leq\left(\int_0^1M^ndx\right)^\frac{1}{n} =M$$
I cannot figure out what to do next.
 A: Let $M$ be as above, note that $M>0$ by assumption. Choose $\epsilon>0$ and let $E_\epsilon = \{x | f(x) >M-\epsilon \}$. Since $f$ is continuous, we have $m E_\epsilon >0$, and so $M \ge \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge (M-\epsilon) \ (m E_\epsilon)^{\frac{1}{n}}$. Hence $\liminf_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge M-\epsilon$. Since $\epsilon$ was arbitrary, we have $\liminf_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} \ge M$, from which it follows that $\lim_n \left(\int_0^1(f(x))^ndx\right)^\frac{1}{n} = M = \max_{x \in [0,1]} f(x)$.
A: Your guess that it should be the maximum is a good guess. You have shown that the limit must be $\leq M$. We will now show that the limit must be greater than or equal to $M-\epsilon$ for any $\epsilon$, from which you can conclude that the limit is indeed $M$.
Since $f(x)$ is continuous, given $\epsilon > 0$, there exists a $\delta$ such that
$$f(x) > M - \epsilon$$ for all $x \in (x_0 -\delta, x_0 + \delta)$. Hence, we have
$$\int_0^1 f(x)^n dx > \int_{x_0 - \delta}^{x_0 + \delta} f(x)^n dx > \int_{x_0 - \delta}^{x_0 + \delta} (M - \epsilon)^n dx = (M-\epsilon)^n 2\delta$$
Hence for any $\epsilon > 0$,
$$\left(\int_0^1 f(x)^n dx\right)^{1/n} > (M-\epsilon)(2 \delta)^{1/n}$$
Letting $n \to \infty$, we get what we want, i.e.,
$$\lim_{n \to \infty}\left(\int_0^1 f(x)^n dx\right)^{1/n} \geq (M-\epsilon)$$
