Proving that $\lim\limits_{x \to 0} x \cdot (\log(x))^\alpha = 0$ for every $\alpha > 0$? 
$\lim_{x \to 0} x \cdot (\log(x))^\alpha = 0$ for every $\alpha > 0$?

It should tend to zero for every $\alpha \in \Bbb R_+$, though I can't find a rigorous way to prove it.
 A: Set $u= x^{1/\alpha} $ then $$  x \cdot \log(x)^\alpha=( x^{1/\alpha} \cdot \log(x))^\alpha = (\alpha x^{1/\alpha} \cdot \log(x^{1/\alpha}))^\alpha  =  (\alpha u \log(u))^\alpha$$
Hence, 
$$\lim_{x \to 0} x \cdot (\log(x))^\alpha =\lim_{u \to 0} (\alpha u \cdot \log u)^\alpha = 0$$
A: Note that the limit is not well defined for $\alpha$ real since $\log x<0$ for $x\to 0^+$, thus assume $\alpha\in\mathbb{N}$.
It is immediate to prove the limit by reiterating l'Hopital, indeed let $x=\frac1y \quad y\to+\infty$
$$x \cdot (\log x)^\alpha=\frac{(-\log y)^{\alpha}}{y}\stackrel{\text{H.R.}}=\frac{-\alpha(-\log y)^{\alpha-1}}{y}...$$
and so on till the exponent for $\log$ is $0$, then
$$...=\frac{c}{y}\to 0$$
As an alternative let $x=e^{-z} \quad z\to+\infty$, thus
$$x \cdot (\log x)^\alpha=\frac{(-z)^{\alpha}}{e^z}\to0$$
which is a standard limit.
A: if you want another really easy way to do it
$x* \log^\alpha x=\frac{(-\log\frac{1}{x})^\alpha}{\frac{1}{x}}=\frac{(-\log u)^\alpha}{u}=\Big(\frac{-\log u}{u^\frac{1}{\alpha}}\Big)^\alpha$
By making the change of variables $u=1/x$ the limits go to infinity instead of zero. Now you apply you only need to apply L'hopital's rule once inside the parentheses. I'm going to clean this up a bit by saying $\alpha=1/f$
$-\frac{1}{u}*(fu^{f-1})^{-1}=-(fu^f)^{-1}$
And thus, for any finite f>0, the limit goes to zero.
