How to compute $\lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} f(t)t \space dt $? For a continuous function $f: R \to R $ Define:
$$ \lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} f(t)t \space dt $$
Since the function is continuous I can assume that it's also integrable since continuity implies integrability. I assume furthermore that there exists a function $F$ which is an antiderivative of $f$ for which the following are true:
$$\int_{a}^{b} f(x) dx =F(b)-F(a) \space\space\space\space\space a,b\in R \space  $$
$$ \lim_{x \to x_{0}} \frac{F(x)-F(x_{0})}{x-x_{0}}=f(x)$$
And that for $f$
$$ \lim_{x \to x_{0}} f(x)=f(x_{0})$$
In order to find the limit, I used partial integration and ended up with:
$$ \lim_{x \to 0} \frac{F(x)(x-1) +F(0)}{x^2}$$
At this point, I tried to use L'Hôpital's rule and ended up with the value $\frac{f(0)}{2}$ which seems totally wrong to me .
Any advice would be appreciated, I mainly think that my solution idea is wrong, but I am stuck.
 A: By L'Hôpital's rule, this limit is equal to
$$\lim_{x\to 0} \frac{xf(x)}{2x} = \frac12\lim_{x\to 0} f(x) = \frac{f(0)}2.$$
(Use the first Fundamental Theorem of Calculus to differentiate the integral, since the integrand is guaranteed continuous.) You were correct.
A: The limit is indeed $f(0)/2$. 
L'Hôpital's Rule applies because the limit of the quotient of the derivatives of numerator and denominator exists. Thus
$$
\lim_{x\to0}\frac1{x^2}\int_0^x f(t)\,t\,dt
=\lim_{x\to0}\frac{x\,f(x)}{2x}=\frac{f(0)}2
$$
A: An approach not relying on L'Hopital's Rule.
We have, for $x\neq 0$ and with the change of variable $u=\frac{t}{x}$, $$
\frac{1}{x^2}\int_0^x t f(t)dt
= \int_0^1 u f(xu) du
$$
Now, it is not hard to prove that
$$
\lim_{x\to 0}\int_0^1 u f(xu) du
= \int_0^1 u \lim_{x\to 0} f(xu) du
= \int_0^1 u f(0) du
= f(0)\left [\frac{x^2}{2}\right]^1_0
= \frac{f(0)}{2}
$$
(where the swapping limit/integral can be justified e.g. by arguing about uniform convergence, using continuity of $f$).
A: Write $f(t) $ as $f(t) - f(0)+f(0)$ and then the desired limit is $$\lim_{x\to 0}\frac{1}{x^2}\int_{0}^{x}\{f(t)-f(0)\}t\,dt+\frac{f(0)}{2}\tag{1}$$ Next we can show that the first limit above is $0$. Let $\epsilon >0$ be given. Then by continuity we have a $\delta>0$ such that $$|f(x) - f(0)|<\epsilon $$ whenever $|x|<\delta$. Thus we have $$\left|\int_{0}^{x}\{f(t)-f(0)\}t\,dt\right |\leq\int_{0}^{x}|f(t)-f(0)|t\,dt<\frac{\epsilon x^2}{2}$$ whenever $0<x<\delta$. Similar inequality holds when $-\delta <x<0$. It thus follows that the first limit in $(1)$ above is $0$. Thus the desired limit is $f(0)/2$.
