How to compute $\lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} f(t)t \space dt $?

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.

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 $$ \lim_{x\to0}\frac1{x^2}\int_0^x f(t)\,t\,dt =\lim_{x\to0}\frac{x\,f(x)}{2x}=\frac{f(0)}2 $$

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 $f$).