The electric field of radius $R$ and a uniform positive surface charge density $\sigma$ at a distance $x$ from its center is given as $$E = \frac{\sigma}{2 \epsilon_0}\left( 1 - \frac{1}{\left(\frac{R^2}{x^2}\right) + 1}\right).$$
I am asked to show that for $x\gg R$, that $E = \frac{Q}{4\pi\epsilon x^2}$.
This is what I've done (but it's wrong):
So since $\sigma$ is the charge density of the disk, $\sigma = \frac{Q}{\pi R^2}$. Substituting this, we get $$E = \frac{Q}{2\pi R^2 \epsilon_0}\left( 1 - \frac{1}{\left(\frac{R^2}{x^2}\right) + 1}\right).$$ Further, as $x \gg R$, then the $\frac{R^2}{x^2}$ term evaluates to 0, so $E$ is therefore 0 at these conditions.
This is wrong, so can someone please explain why the answer is $E = \frac{Q}{4\pi\epsilon x^2}$?