1. ## Limit convergence?

I need to show that

$\lim_{n}(\sqrt{n+1}-\sqrt{n})=0$

and

$\lim_{n}(\sqrt{n^2-n}-n)=-\frac{1}{2}$

Help would be greatly appreciated. Thanks.

2. Originally Posted by miatchguy
I need to show that

$\lim_{n}(\sqrt{n+1}-\sqrt{n})=0$
The standard trick here is to multiply with $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$:

$\lim_{n}(\sqrt{n+1}-\sqrt{n})=\lim_{n\to\infty}\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to\infty}\frac{1}{ \sqrt{n+1}+\sqrt{n}}=0$

and

$\lim_{n}(\sqrt{n^2-n}-n)=-\frac{1}{2}$
Well, let's try the same basic trick once more

$\lim_{n\to\infty}(\sqrt{n^2-n}-n)=\lim_{n\to\infty}\frac{n^2-n-n^2}{\sqrt{n^2-n}+n}=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}$
$=\lim_{n\to\infty}\frac{-1}{\sqrt{1-\frac{1}{n}}+1}=-\frac{1}{2}$

3. Originally Posted by Failure
The standard trick here is to multiply with $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$:

$\lim_{n}(\sqrt{n+1}-\sqrt{n})=\lim_{n\to\infty}\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to\infty}\frac{1}{ \sqrt{n+1}+\sqrt{n}}=0$

Well, let's try the same basic trick once more

$\lim_{n\to\infty}(\sqrt{n^2-n}-n)=\lim_{n\to\infty}\frac{n^2-n-n^2}{\sqrt{n^2-n}+n}=\lim_{n\to\infty}\frac{-n}{\sqrt{n^2-n}+n}$
$=\lim_{n\to\infty}\frac{-1}{\sqrt{1-\frac{1}{n}}+1}=-\frac{1}{2}$
Thanks alot for the help there. My professor was showing me a method based on graphing, where I'd use the quotient the slopes to bound the difference of the values. It got to be very tricky. I don't know why I didn't see this. Thanks again.