Evaluating $ \lim_{ x \to 1} \frac{\int_{1}^{x}\ln^2(t) dt}{(x-1)^3} $
When applying l'hospital, why didn't you derive $\left(x-1\right)^3$, that'd make de $1/3$ appear ?
It'd give
$$ \lim\limits_{x \rightarrow 1}\frac{\ln^2\left(x\right)}{3\left(x-1\right)^2} $$