What is $\lim_{\alpha\rightarrow0} \left(\alpha\log^2{\alpha}\right)$? What is the limit of $\alpha\log^2{\alpha}$ as $\alpha\rightarrow0$?, these are my workings so far:
\begin{align}
\lim_{\alpha\rightarrow0} \left(\alpha\log^2{\alpha}\right)
%
&= 
%
\lim_{\alpha\rightarrow0} \left(\frac{\alpha}{\log^{-2}{\alpha}}\right)\\
%
&= 
%
\lim_{\alpha\rightarrow0} 
\left(
-\frac{1}{2}
\left\{
\frac{\alpha}{\log^{-3}{\alpha}}
\right\}
\right)
\\
%
&= 
%
\lim_{\alpha\rightarrow0} 
\left(
\frac{1}{6}
\left\{
\frac{\alpha}{\log^{-4}{\alpha}}
\right\}
\right)
\\
%
&= \cdots \\
%
%
&= 
%
\lim_{\alpha\rightarrow0} 
\left(
\lim_{n\rightarrow\infty}
\left[
\frac{\left(-1\right)^n}{n!}
\left\{
\frac{\alpha}{\log^{-(n+1)}\alpha}
\right\}
\right]
\right)\\
%
%
&= 
%
\lim_{\alpha\rightarrow0} 
\left(
\lim_{n\rightarrow\infty}
\left[
\frac{\left(-\alpha\right)^n}{n!}
\log^{(n+1)}\alpha
\right]
\right)
\end{align}
However I am now slightly stuck!
EDIT: Oh jeez, for a fixed $\alpha>0$
\begin{align}
\lim_{n\rightarrow\infty}
\left[
\frac{\left(-\alpha\right)^n}{n!}
\log^{(n+1)}\alpha
\right] = 0
\end{align}
doesn't it?
 A: 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.
A: 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 $\lim\limits_{\beta\to0}\beta\log\beta=0$.
A: $f(x)=x\ln(x)^2$
$f'(x)=\ln(x)\left(\ln(x)+2\right)\quad$ 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.
A: Write $a \log^2a=\frac{\log^2a}{ \frac{1}{a}}$ and aply  L'Hospital's rule two times (doing the same trick before you differentaite the second time.
