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.
 A: 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 $n>0$:
$$\lim_{x \to +\infty} \frac{x^n}{\log x} = +\infty$$
A: $\lim_{x \to +\infty}(\sqrt{x}-\log(x))=[\infty-\infty]=\lim_{x \to +\infty}\frac{(x-\log(x)^2)}{(\sqrt{x}+\log(x))}=\lim_{x \to +\infty}\frac{\sqrt x-\frac{log(x)^2}{\sqrt x}}{\frac{\log x}{\sqrt x}+1}$
$-\frac{\log(x)^2}{\sqrt x}$ and $\frac{\log(x)}{\sqrt x}$ both tend to zero as $x$ tends to $\infty$.
Then your limit is $\lim_{x \to \infty}\sqrt x=+\infty$.
A: $$\lim_{x \to +\infty}(\sqrt{x}-\log x)$$
$$\lim_{x \to +\infty}
   \left(\dfrac{\sqrt{x}-\log x}{1} 
   \cdot 
   \dfrac{\sqrt{x}+\log x}{\sqrt{x}+\log x}
   \right)$$
$$\lim_{x \to +\infty}
   \left( 
   \dfrac{x-(\log x)^2}{\sqrt{x}+\log x}
   \right)$$
$$\lim_{x \to +\infty}
   \left( 
   \dfrac{1-\frac{2\log x}{x}}{\frac{1}{2\sqrt x}+\frac 1x}
   \right) \to \dfrac 10\to \infty$$
