Q:

$\lim_{x\to 0} x^{\alpha}e^{|\log x|^{1/2}}=0$

Given any $\alpha > 0$, I need to show that for $ x \in [0,\infty)$
 \begin{equation}
     \lim_{x\to 0} x^{\alpha}e^{|\log x|^{1/2}}=0
\end{equation}
I have tried using L'Hospital's rule. But I am not able to arrive at answer. 
Thank you in advance.

A:

Let
$f(x)
=x^{a}e^{|\log x|^{1/2}}
$.
$\ln f(x)
=a\ln x+|\log x|^{1/2}
$.
Let
$x = 1/y$,
so
$y \to \infty$
as
$x \to 0$.
$\ln f(1/y)
=a\ln (1/y)+|\log (1/y)|^{1/2}
=-a\ln (y)+|\log y|^{1/2}
$.
The key is that
$\dfrac{|\log y|^{1/2}}{\ln(y)}
\to 0$
as $y \to \infty$.
Therefore
$\ln f(1/y)
=\ln (y)(-a +\dfrac{|\log y|^{1/2}}{\ln(y)})
$.
Since
$\dfrac{|\log y|^{1/2}}{\ln(y)}
\to 0$
as $y \to \infty$
and $a > 0$,
$-a +\dfrac{|\log y|^{1/2}}{\ln(y)}
\lt -a/2$
for large enough $y$
so that
$\ln f(1/y)
\to -\infty$
so
$f(1/y) \to 0$.
Note that this works
for any exponent
less than $1$,
not just
$\frac12$.

