Find simple limit with basic methods

I need to find simple limit: $$\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}$$

I can't use L'Hôpital's rule, which I think is natural for such problems. How to calculate such limit using only basic methods (definition and simple transformations etc.)?

It's one problem (from 50 problems) I don't know how to solve from my homework.

Thanks for help.

-

If you really mean what's on the post, you can show the limit diverges to $+\infty$. I'm pretty sure you mean $$\lim \;\sqrt{n+1}-\sqrt n$$

Then note that

$$\sqrt{n+1}-\sqrt n=\frac 1 {\sqrt{n+1}+\sqrt n}$$

The rest is pretty straightforward.

-
thanks, i edited my question :). –  golert Dec 6 '12 at 23:56

$$\lim_{n\to\infty}\left(\frac{(\sqrt{n+1}-\sqrt{n})\cdot(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}\right)$$
$$\lim_{n\to\infty}\left(\frac{n+1-n}{(\sqrt{n+1}+\sqrt{n})}\right)$$
Try conjugate multiplication - $\sqrt{n+1}+\sqrt n$. I assume, that you mean $\sqrt{n+1}-\sqrt n$, otherwise it's $\infty$.