Neat result about the average integral of a converging function Let $f:\mathbb{R_+}\to\mathbb{R_+}$ such that $\lim\limits_{x\to+\infty} f(x)\int\limits_0^x f(t) \, \mathrm{d}t = 1$.
Prove that $f(x) \underset{x\to+\infty}{\sim} \frac{1}{\sqrt{2x}}$.

The previous question has us prove that if $f$ converges to 1 then its integral is equivalent to $x$. I've guessed we are to use this result on the new function $f(x)\int\limits_0^x f(t) \, \mathrm{d}t$, but I haven't been able to go any further than this:
$$\int\limits_0^x f(t) \int\limits_0^t f(u) \, \mathrm{d}u \, \mathrm{d}t \underset{x\to+\infty}{\sim} x$$
What's the next step?
 A: let $\int_{0}^{x}f(t)dt$ = F(x) So $\forall$ $\epsilon$ $>$ 0 $\exists$ $A$ such that $\forall$ x $\geq$ $A$ $\Rightarrow$  -$\epsilon$ $<$ $f(x)$ $F(z)$ - $1$ <$\epsilon$ then 2($1-\epsilon$)($x-A$)<$F^{2}(x)-F^{2}(A)$<2($1+\epsilon$)($x-A$)
then we can show that $\lim_{x \to \infty}$ $\frac{F(x)}{\sqrt{2x}}$ = 1 and because $\lim_{x \to \infty}$$f(x)F(x)$$=1$ we are done.
A: The key here is L'Hospital's Rule. Let $g(x) =\int_{0}^{x}f(t)\,dt$. Then we are given that $f(x) g(x) \to 1$ as $x\to\infty$. Or $g(x) g'(x) \to 1$ so that $((g(x)) ^{2})'\to 2$. Via L'Hospital's Rule we can see that $$\lim_{x\to\infty} \frac{(g(x)) ^{2}}{x}=2$$ and since $g$ is positive we can see that $g(x) /\sqrt{x} \to\sqrt{2}$. Therefore $$\lim_{x\to\infty} \sqrt{2x}f(x)=\sqrt{2}\lim_{x\to\infty}\frac{\sqrt{x}}{g(x)}\cdot f(x) g(x) =\sqrt{2}\cdot\frac{1}{\sqrt{2}}\cdot 1=1$$ And this is what we mean by $f(x) \sim 1/\sqrt{2x}$ as $x\to\infty$. Note that the above assumes continuity of $f$ so that the relation $g'=f$ holds. 
A: (I'm self-answering as no other answer was quite as lightweight as what Joey's answer inspired me.)

Let $F(x)=\int_0^xf(t)dt$, and $g(x)=f(x)F(x)$. We have $g(x)\sim1$ and therefore we can write:
$$\int_0^xg(t)dt\sim1 \iff \frac{1}{2}F^2(x)\sim x$$
And:
$$f(x)F(x)\sim1 \implies f^2(x)2x\sim1 \implies f(x) \sim \frac{1}{\sqrt{2x}}$$
The last implication relies on $(f(x)\sqrt{2x})^2$ converging to $1$.
We thus have the required result only using equivalences, without relying on L'Hospital's Rule or the definition of a limit.
A: Let $F(x)=\int_0^xf(t)dt$,Then, 
$$\lim_{x\to\infty}f(x)F(x) = 1\Longleftrightarrow \lim_{x\to\infty}\frac{d}{dx}\left( \frac{1}{2}F^2(x)\right)=1. $$

Claim
   $$ \lim_{x\to\infty}\frac{d}{dx}\left( \frac{1}{2}F^2(x)\right)=1 \implies F(x)\sim \sqrt{2x} $$ see the proof below

Therefore, as $x\to \infty $ we have, 
$$ F(x)\sim \sqrt{2x}\Longleftrightarrow \frac{1}{F(x)}\sim 
\frac{1}{\sqrt{2x}}\qquad x\to \infty  $$
And by assumption, as $x\to \infty $
$$ f(x)F(x)\sim1 \Longleftrightarrow f(x)\sim \frac{1}{F(x)}\sim \frac{1}{\sqrt{2x}} $$
that is $$f(x)\sim \frac{1}{\sqrt{2x}}.$$
Proof of The claim
For $\varepsilon = 1$ there exists $\delta>0$ such that for all $x>\delta$ we have 
$$\left|\frac{d}{dx}\left( \frac{1}{2}F^2(x)\right)-1\right|<1 \Longleftrightarrow 1<\frac{d}{dx}\left( \frac{1}{2}F^2(x)\right)<3$$
By integration we have that for all $x>\delta,$
$$  (x-\delta) +\frac{1}{2}F^2(\delta)< \frac{1}{2}F^2(x)<3(x-\delta) +\frac{1}{2}F^2(\delta)$$
That is
 $$F^2(x)\sim x \Longleftrightarrow F(x)\sim \sqrt{2x}$$
 We have $g(x)\sim1$ and therefore we can write:
Since for $x\to \infty$
$$ (x-\delta) +\frac{1}{2}F^2(\delta)\sim x\qquad and \quad 3(x-\delta) +\frac{1}{2}F^2(\delta)\sim x $$
