Shortcut for computing $ \lim_{x \to 0^+} \frac{1}{2x}\int_0^x \ln(t)t^2\,dt $

Suppose we want to find $$ \lim_{x \to 0^+} \frac{1}{2x}\int_0^x \ln(t)t^2\,dt $$ without computing the integral. What I would do is this :

We know that $ 0 \leq t^2 \leq x^2$ for all $t \in (0,x)$

Then $ \ln(t)x^2 \leq \ln(t)t^2 \leq 0$ since $ \ln(t) \lt 0 ,x \in (0,1)$

By monotonicity of integration,

$$\begin{align*}x^2\int_0^x\ln(t)\,dt &\leq \int_0^x\ln(t)t^2\,dt \leq0\\x^2(x\ln(x)-x) &\leq \int_0^x\ln(t)t^2\,dt \leq0\end{align*}$$When we take the limits at the extremes of the inequality, both expressions are zero. By squeeze theorem, the original limit is zero.

I have certain doubts regarding this solution, especially because I am using monotonicity of integrals based in an inequality on an open Interval, with the integral on the left being improper.