Approximation of $\frac{x}{\sqrt{x^2+R^2}}$ How do you prove this statement?

If $x\gg R$ then
  $$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$

I have no ideas even how to start.

Basically, this can be seen by a Taylor series expansion around $0$. Note that $x\gg R$ "means" $\frac{R}{x} \approx 0$: so you would need to make this quantity appear, first.
$$\begin{align}
\frac{x}{\sqrt{x^2+R^2}} &= \frac{x}{x\sqrt{1+\left(\frac{R}{x}\right)^2}}
= \frac{x}{x}\left(1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\left(\frac{R}{x}\right)^2\right)\right) \\
&= 1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\frac{R^2}{x^2}\right)
\end{align}$$
using the fact that $(1+u)^\alpha = 1+\alpha u + o(u)$ when $u\to0$ (for any fixed $\alpha \in\mathbb{R}$).
