Why ${(0^+)}^{+ \infty}$ is well defined and not ${(0^+)}^{- \infty}$? I want to compute the limit of $|x|^{\frac{1}{x}}$ when $x$ tends to $0$.
When $x$ tends to $0$ from above, I get ${(0^+)}^{\frac{1}{0^+}}= {(0^+)}^{+\infty}=0$. However when, I approach $0$ from below I get $ {(0^+)}^{\frac{1}{0^-}} = {(0^+)}^{-\infty}$ and I can't give a value for this. ( why ?)
However, I was able to find a value for this limit by going through the exponential definition of $x$ as follow :
$$\lim_{x \to 0^-} |x|^{\frac{1}{x}} =\lim_{x \to 0^-} e^{\frac{1}{x} \ln(|x|)}=\lim_{x \to 0^-} e^{\frac{\ln(|x|)}{x} } = e^{\lim_{x \to 0^-}\frac{\ln(|x|)}{x} }= e^{\frac{-\infty}{0^-}}=e^{+\infty}=+\infty$$
So, my question is, why am I forced to go through this definition? I might have been able to give a value to ${(0^+)}^{- \infty}$

"show me how to formalize my expressions please". Maybe you mean like this?
Assume $a_n\to 0^+$ and $b_n\to -\infty$ for $n\to \infty$. Note that $a_n>0$ for large enough $n$, so $a_n^{b_n}$ is well-defined large enough $n$. Then $a_n^{b_n}\to \infty$. And you basically did a proof yourself, starting with
$$
a_n^{b_n}
= e^{b_n\log a_n}.
$$
Now, $\log a_n \to -\infty$ by continuity, so $b_n\log a_n\to+\infty$, which gives
$$
a_n^{b_n}
= e^{b_n\log a_n}
\to +\infty.
$$
In short, $(0^+)^{-\infty} = +\infty$.
Prerequisites for this proof are
$
(-\infty)\cdot(-\infty) = +\infty$ and $c^{+\infty} = +\infty,
$
where $c>1$ is a constant greater than $1$.
