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.

$\lim_{x \to +\infty}(\sqrt{x}-\log(x))=[\infty-\infty]=\lim_{x \to +\infty}\frac{(x-\log(x)^2)}{(\sqrt{x}+\log(x))}=\lim_{x \to +\infty}\frac{\sqrt x-\frac{log(x)^2}{\sqrt x}}{\frac{\log x}{\sqrt x}+1}$
$-\frac{\log(x)^2}{\sqrt x}$ and $\frac{\log(x)}{\sqrt x}$ both tend to zero as $x$ tends to $\infty$.
Then your limit is $\lim_{x \to \infty}\sqrt x=+\infty$.
