How to solve this limit ( No l'Hôpitale)

Question

Hi I'm sorry for bothering you a lot, but i'm practising for my last test When ever i try to do something it's inverting to NaN situation
Can any one help me
Thanks a lot
Note : Obviously It's not 2sided limit (0+)(0-)
With L'Hôpitale I've found that

right side limit is +∞

left side limit is -∞

$$ \bbox[] { \lim_{x\to0} \left( \frac{e^x-x^x}{x^2} \right) \ } $$

Answer 1

Take $x\to 0^+$ and rewrite $x^x=e^{x\ln x}$. Thus $$ \frac{e^x-x^x}{x^2}=\frac{1}{x}\cdot \bigg[\frac{e^x-1}{x}-\frac{e^{x\ln x}-1}{x\ln x}\cdot\ln x\bigg]\to +\infty\cdot[1-1\cdot (-\infty)]=+\infty. $$

Answer 2

HINT:

For $x>0$, we have $x^x=1+x\log(x)+O(x\log(x))^2$.

And $\lim_{x\to 0^+}\frac{1-\log(x)}{x}=\infty$