Prove that exists $\lim_{x\rightarrow \infty }\frac{f(x)}{x}$ and determine its value Let $f:[0,\infty )\rightarrow \mathbb{R}$ be an increasing function, such that $\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=1$.
Prove that exists $\lim_{x\rightarrow \infty }\frac{f(x)}{x}$ and calculate this limit.

If $f$ would be continuous, we'd have $1=\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=\lim_{x\rightarrow \infty }\frac{f(x)}{2x}$, therefore $\lim_{x\rightarrow \infty }\frac{f(x)}{x}=2.$
But we don't know if $f$ is continuous or not, and I wasn't able to find any other idea.
 A: Suppose $x=n$ is an integer. By the Stolz-Cesaro theorem,
\begin{eqnarray}
1&=&\lim_{x\rightarrow \infty }\frac{\int_0^xf(t)dt}{x^2}=\lim_{n\rightarrow \infty }\frac{\int_0^nf(t)dt}{n^2}=\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{(n+1)^2-n^2}\\
&=&\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{2n+1}=\frac12\lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{n}
\end{eqnarray}
and hence
$$ \lim_{n\rightarrow \infty }\frac{\int_n^{n+1}f(t)dt}{n}=2 $$
Since $f(t)$ is increasing, one has
$$ f(n)\le\int_n^{n+1}f(t)dt\le f(n+1) $$
and hence
$$ \frac{f(n)}{n}\le\frac{\int_n^{n+1}f(t)dt}{n}\le \frac{f(n+1)}{n+1}\frac{n+1}{n}\le \frac{f(n+1)}{n+1}. $$
Letting $n\to\infty$, one obtains
$$ \lim_{n\rightarrow \infty }\frac{f(n)}{n}=\lim_{n\rightarrow \infty }\frac{\int_{n}^{n+1}f(t)dt}{n}=2.$$
For general $x>0$, let $\lfloor x\rfloor=n$ and then
$$ n\le x<n+1.$$
Since $f(t)$ is increasing, one has
$$ f(n)\le f(x)\le f(n+1) $$
and hence
$$ \frac{f(n)}{n+1}\le \frac{f(x)}{x}\le \frac{f(n+1)}{x}\le\frac{f(n+1)}{n}. $$ 
By the Squeeze Theorem, one has
$$ \lim_{x\to\infty}\frac{f(x)}{x}=\lim_{n\to\infty}\frac{f(n)}{n}=2.$$
