Show that $\lim _{x\downarrow 0}\ x^{\alpha }\log \left(x\right) = 0$ for $\alpha > 0$

Question

How can one show that

$$\lim _{x\downarrow 0}\ x^{\alpha }\log \left(x\right) = 0$$

for $\alpha > 0$?

I know that the logarithm function is slower than any power, hence it loses when coupled to powers, meaning that $x^\alpha \log (x) \to 0$ as $x \to 0^+$, but that's just a repetition of what's given...

Answer 1

Let $\alpha>0$ $$L=\lim_{x \rightarrow 0^+} x^{\alpha} \log x \rightarrow 0\times \infty$$ By L'Hospital Rule, we get $$L=\lim_{x\rightarrow 0^+} \frac{\log x}{x^{-\alpha}}= \lim_{x\rightarrow 0^+} \frac{\log x}{x^{-\alpha}}= \lim_{x \rightarrow 0} \frac{1/x}{-\alpha x^{-\alpha-1}}= \lim_{x \rightarrow 0} -\frac{x^{\alpha}}{\alpha}=0$$

Answer 2

If $y = \dfrac1{x}$ then

$\begin{array}\\ x^a\log(x) &=\dfrac1{y^a}\log(\dfrac1{y})\\ &=-\dfrac{\log(y)}{y^a}\\ \end{array} $

and since $\lim_{y \to \infty} \dfrac{\log(y)}{y^a} =0 $ for any $a > 0$, $\lim_{x \to 0} x^a \log(x) =0 $ for any $a > 0$.