$\lim_{x \to 0}|x|^{\frac{1}{x^2}}$?

Question

*I can't use LHopital.

Calculate the limit, if it exists:

$$\displaystyle \lim_{x \to 0}|x|^{\frac{1}{x^2}}$$

My attempt

$$ \lim_{x \to 0^+}|x|^{\frac{1}{x^2}} = \exp\left( \lim_{x \to 0}\ln|x|^{\frac{1}{x^2}}\right) = \exp\left( \lim_{x \to 0}\frac{1}{x^2}\ln|x|\right) = \exp\left( \lim_{x \to 0^+}\frac{\ln x}{x^2}\right).$$

And here i am stuck.

Answer 1

We have

$\lim_{x\rightarrow 0}\frac{1}{x^2}=+\infty$ and $\lim_{x\rightarrow 0} \ln |x| = -\infty$

So the product $\lim_{x\rightarrow 0}\frac{\ln|x|}{x^2}=-\infty$

Hence the limit is 0.