What is $\lim_{x \to 0} \frac{\ln(x^{{10}^8})}{x^2}$? What does this evalute to?

$$\lim_{x \to 0} \frac{\ln(x^{{10}^8})}{x^2}$$

I know basic facts relating to limits of $\ln$ and $x$, and WolframAlpha says the answer is $-\inf$. I can convince myself that this is true, but how can one show it?

Thanks.
You have $\lim_{x \to 0^{+}} \log x = -\infty$ and $\lim_{x\to 0^{+}} \frac{1}{x^2} = \infty$ so $\lim_{x \to 0^{+}} \frac{\log x}{x^2} = -\infty$. 

What this is really saying is that $\log x$ is unbounded below as $x\to 0^{+}$ and $1/x^2$ is unbounded above as $x\to 0^{+}$ so their product is also unbounded below as $x\to 0^{+}$.  

Edit: as per Jonas's comment, you can replace all instances of $\lim_{x\to 0^{+}} \log x$ with $\lim_{x\to 0} \log |x|$ since $\log (x^{10^8}) = 10^8 \log  |x|$.
Note $\mathrm{ln}(x^{10^8}) = 10^8\mathrm{ln}(x)$ so ignore the $10^8$ for the moment and just use L'Hospital's rule on $\frac{\mathrm{ln}(x)}{x^2}$.

Edit: Oops, I misread the limit as going to $\infty$.  In fact the limit is going to $0$, in which case both $-\mathrm{ln}(x^{10^8})$ and $\frac{1}{x^2}$ go to $\infty$, so their product goes to $\infty$, so $\frac{\mathrm{ln}(x^{10^8})}{x^2}$ goes to $-\infty$.