Solution

Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 11D / a / Solution

{title2=$\sqrt u-\sqrt v=(u-v)/(\sqrt u+\sqrt v)$}

For $u,v\ge0$ with $u+v>0$, multiplication by the sum of the <square roots> gives
$$
\sqrt u-\sqrt v=\frac{u-v}{\sqrt u+\sqrt v}.
$$
This identity exposes small differences hidden by cancellation. For example, $\sqrt{n^2+n}-n=1/(\sqrt{1+1/n}+1)\to1/2$. It also converts normalized square-root increments into terms comparable with the <harmonic series>.

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 Incoming links: Rationalization of a square-root difference