Prove that the limit of $\sqrt{n+1}-\sqrt{n}$ is zero How would I go about proving that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$? I have tried to use Squeeze theorem but have not been able to come up with bounds that converge to zero. Additionally, I don't think that converting to polar is possible here.
 A: Note that for positive $n$ we have
$$\sqrt{n}\lt \sqrt{n+1}\lt \sqrt{n}+\frac{1}{2\sqrt{n}}.\tag{1}$$
The second inequality in (1) holds because 
$$\left(\sqrt{n}+\frac{1}{2\sqrt{n}}\right)^2=n+1+\frac{1}{4n}\gt n+1.$$
It follows from (1) that
$$0\lt \sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}.$$
Now Squeeze. 
A: $$ \sqrt{n+1}-\sqrt{n}
 = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}
 = \frac{1}{\sqrt{n+1}+\sqrt{n}}
 < \frac{1}{2\sqrt{n}}
$$
A: Using conjugate multiplication can be quite useful in cases like that:
$$\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n}){\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}}={(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})\over \sqrt{n+1}+\sqrt{n}}={(n+1)-n\over \sqrt{n+1}+\sqrt{n}}={1\over \sqrt{n+1}+\sqrt{n}}$$.
The last result is pretty easy to work with.
A: One way is by using the mean value theorem. Specifically, let $f(x) = \sqrt x$. Then, for each $x > 0$, we know that $\displaystyle f(x+1) - f(x) = \frac{f(x+1) - f(x)}{(x+1) - x} = f'(c)$ for some $c$ in the interval $(x, x+1)$. Since $\displaystyle f'(x) = \frac1{2\sqrt x}$ is strictly decreasing we conclude that $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.
