Calculate the limit using de L'Hopital's rule Calculate the following limit: 
$\lim_{x \to +\infty}(\sqrt{x}-\log x)$
I started like this:
$\lim_{x \to +\infty}(\sqrt{x}-\log x)=[\infty-\infty]=\lim_{x \to +\infty}\frac{(x-(\log x)^2)}{(\sqrt{x}+\log x)}=$
but that's not a good way...
I would be gratefull for any tips.

Hint
If you have to use l'Hôpital; this limit is easier to find (*):
$$\lim_{x \to +\infty} \frac{\sqrt{x}}{\log x} 
= \lim_{x \to +\infty} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}} =\lim_{x \to +\infty}\frac{\sqrt{x}}{2} = +\infty$$
Can you see how this would help for your limit as well?
If not (hoover over), rewrite:

 $$\sqrt{x}-\log x = \left( \frac{\sqrt{x}}{\log x} - 1 \right) \log x$$


(*) With a similar calculation, it's easy to show and worth remembering that for $n>0$:
$$\lim_{x \to +\infty} \frac{x^n}{\log x} = +\infty$$
