Limit of $\sqrt n (\sqrt{n+1}-\sqrt n)$ How should I go about calculating $$\lim_{n \to \infty} \sqrt n (\sqrt{n+1}-\sqrt n)$$
I found empirically it approaches $\frac 1 2$ (and fast), but it's been decades since I last took a formal calculus class, and I don't know how to approach this.
 A: Hint.
$$\sqrt n \left(\sqrt{n+1}-\sqrt n\right) = \frac{\sqrt n}{\sqrt{n+1} + \sqrt n} = \frac{1}{\sqrt{1 + \frac1n} + 1}$$
A: Make use of
$$\sqrt n (\sqrt {n+1}-\sqrt n)=\sqrt n (\sqrt {n+1}-\sqrt n)\frac{ \sqrt {n+1}+\sqrt n}{\sqrt {n+1}+\sqrt n}$$
This simplifies to
$$\sqrt n\frac{ (\sqrt n+1)^{2}-\sqrt n^{2}}{\sqrt n+1)+\sqrt n}=\sqrt n\frac{n+1-n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt {n+1}+\sqrt n}=\frac{1}{\sqrt \frac{n+1}{n}+1}$$
This can be taken to the limit since $\lim_{n\rightarrow \infty}\frac{n+1}{n}=1$. There are several alternatives for this direct path. $\lim_{n\rightarrow \infty}\frac{n+1}{n}=\lim_{n\rightarrow \infty}1+ \frac{1}{n}=1$. This is calulating with limits. The denominator can not be developed for example into a Taylor series since  one is not small.
So a substitution is well working: $\frac{n+1}{n}=\left(\frac{m+1}{m}\right)^{2}$. This takes some time to write down. Then the square root is gone and the limit again can be shown to be $\frac{1}{2}$.
$$\lim_{m\rightarrow \infty}\frac{1}{2+\frac{1}{m}}=\frac{1}{2}$$
