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.

Compare these two functions at $x=\infty$. If their quotient $f/g$ tends to $1$, it means $f\sim g$:
$$\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{1-\frac{1}{2}\left(\frac{R}{x}\right)^2}=
\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{\frac{2x^2-R^2}{2x^2}}=
\lim_{x\to\infty} \frac{x\cdot 2x^2}{\sqrt{x^2+R^2}(2x^2-R^2)}=
\lim_{x\to\infty} \frac{2}{\sqrt{1+R^2/x^2}(2-R^2/x^2)}=1$$
