What is $\lim_{\alpha\rightarrow0} \left(\alpha\log^2{\alpha}\right)$?

If you know that:

$$ \forall \alpha,\beta \in ]0,+\infty[, \, \lim \limits_{x \to -\infty} e^{\alpha x}\vert x \vert^{\beta} = 0 \tag{1}$$

Then, it follows that:

$$ \forall \alpha,\beta \in ]0,+\infty[, \, \lim \limits_{\substack{x \to 0 \\ x > 0}} x^{\alpha} \vert \ln(x) \vert^{\beta} = 0. $$

All you need to prove is $(1)$. This is the same as Equation $(2)$ in this post.

Since $\alpha>0$, you can set $\alpha=\beta^2$, so your limit becomes $$ \lim_{\beta\to0}4(\beta\log\beta)^2 $$ and it's known that $\lim\limits_{\beta\to0}\beta\log\beta=0$.

$f(x)=x\ln(x)^2$

$f'(x)=\ln(x)\left(\ln(x)+2\right)\quad$ near $0$ we have $\ln(x)\to-\infty$ so $f'>0$.

Since near zero $f\ \nearrow\ $ positive and continuous, it admits a limit $\lim\limits_{x\to 0^+}f(x)=\ell\ge 0$.

But $\underbrace{f(x)}_{\to \ell}=x\ln(\sqrt{x}^2)^2=x(2\ln(\sqrt{x}))^2=4\sqrt{x}(\sqrt{x}\ln(\sqrt{x})^2)=\underbrace{4\sqrt{x}}_{\to 0}\ \underbrace{f(\sqrt{x})}_{\to \ell}\to 0$

Then $\ell=0$ is forced.