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?)
 A: 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 $$ \lim_{x\to0^+}|x|^{1/x}=0,\qquad \lim_{x\to0^-}|x|^{1/x}=+\infty.$$
A: Try this.
First we have:
$$|x|^{\frac{1}{x}}$$
If we split that up into it's definition which is
$$|x|^{\frac{1}{x}} =
\begin{cases}
x^{\frac{1}{x}},  & \text{if $x > 0$} \\
-x^{\frac{1}{x}}, & \text{if $x < 0$}  \\
\end{cases}$$
So from here, you can find the first limit and second limit separately and see if they match. So for the first limit we have
\begin{align}
&\lim_{x\to0^+} x^{\frac{1}{x}} \\
=&\lim_{x\to0^+} e^{{\ln (x^{\frac{1}{x}})}} \\
=&\lim_{x\to0^+} e^{\frac{1}{x} \cdot {\ln (x)}} \text{ *}\\
=& 0 \bullet
\end{align}
$\bullet$. Okay so, in order to figure out the above limit you convert it to base $e$ which will allow you to bring down the $\frac{1}{x}$ and make your life much easier. Once we have done that, we need to evaluate the limit for $\frac{\ln x}{x}$. We can do this separately by finding the limit for $\frac{1}{x} * \ln x$ as $x$ tends to $0^{+}$ You'll notice that for $\ln x$ it's $-\infty$ and for $\frac{1}{x}$ is $+\infty$ but here, you have to realize that $\frac{1}{x}$ is not growing as fast as $\ln x$ and thus, it won't have much effect on $\ln x$ going to $-\infty$. Therefore, the limit of $\frac{\ln x}{x}$ must be $-\infty$. Also, you cannot take this limit from the left side because $\ln x$ is undefined for negative numbers. Now once, we have found that the limit of those terms is $-\infty$, we must find it for the whole function. Now ask yourself what happens when
$$\lim_{x\to0^+} e^{-\infty}$$
If you look at a graph, you'll see that the answer is $0$.
Now onto looking at the limit from the left side. 
$$\begin{align}
&\lim_{x\to0^-} -x^{\frac{1}{x}} \\
=&\lim_{x\to0^-} -e^{\ln (x^\frac{1}{x})} \\
=&\lim_{x\to0^-} -e^{\frac{1}{x}*\ln (x)} \\
=&\lim_{x\to0^-} -e^{-\infty} \\
=&-\infty
\end{align}$$
Here the procedure is the same but the only difference is that we have a negative sign in front of the $e$ which means the graph is flipped. Again, refer to graph of a simple $e^{x}$ and $e^{-x}$ and it will make a lot more intuitive sense. 
So clearly, both of the limits don't match and as a result, there is no limit for the function. If you need even further clarification, please let me know. 
(*) We can do that step due the laws of logarithms.
