# Prove that exists $\lim_{x\rightarrow \infty }\frac{f(x)}{x}$ and determine its value [duplicate]

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.

## marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 29 '17 at 3:04

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 $$\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{n\to\infty}\frac{f(n)}{n}=2.$$
• The Stolz-Cesaro test says the INVERSE! If $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ exists, then so does $$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}$$ and they are equal. Not the other way around! – Yiorgos S. Smyrlis Sep 28 '17 at 18:38
• The link you gave says exactly the same thing as the comment from @YiorgosS.Smyrlis. What is needed here is to show that the ratio $f(x) /x$ tends to a limit. – Paramanand Singh Sep 28 '17 at 19:36