Difference between following sqrts

Hello everyone

I was wondering if there was a proof for this theorem:

Be [math]\Delta x = \sqrt{x}-\sqrt{x-1}[/math]

then [math]\lim_{x\to\infty}{\Delta x}=0[/math]

Thanks!

Oh wait... Would this be a plausible proof:

[math]\lim_{x\to\infty}{\left(\sqrt{x}-\sqrt{x-1}\right)}[/math]

[math]=\lim_{x\to\infty}{\left[\frac{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}{\sqrt{x}+\sqrt{x-1}}\right]}[/math]

[math]=\lim_{x\to\infty}{\left[\frac{x-(x-1)}{\sqrt{x}+\sqrt{x-1}}\right]}[/math]

[math]=\lim_{x\to\infty}{\left[\frac{1}{\sqrt{x}+\sqrt{x-1}}\right]}\left(=\frac{1}{\infty}\right)=0[/math]

Edited January 19, 201411 yr by Function