Finding the limit of $|x|^{1/x}$ as $x \rightarrow 0$. I have to find the limit of $|x|^{1/x}$ as $x\rightarrow 0$ 
I can't use de l'Hopital's method and derivatives.
Now the limit itself does not exist but it have to be divided into two side the left and right. Then it have to be calculated separately. The right side is 0, the problem is in the left side. Someone managed to solve the limit giving the correct result that is -infinity but I can't figure it out. Basing on my calculus I've arrived to +infinity and if you got to see in wolfram alpha putting lim (-x)^(1/x) as x->0- (the left side) you can see that the limit goes to +infinity. In my opinion the book's solution is wrong, however can someone show me that I'm right (to have a confirm of my solution?)

By definition, $|x|^{1/x}=\exp\frac{\ln|x|}x$. Since $\ln|x|\to-\infty$ and $\frac1x\to\pm\infty$ as $x\to 0^\pm$, the limit as $x\to 0$ does not exist: We have $$ \lim_{x\to0^+}|x|^{1/x}=0,\qquad \lim_{x\to0^-}|x|^{1/x}=+\infty.$$
