Reducing the radius of an electrically charged disc to a point Consider a disc of radius $R$ with electric charge $Q$ evenly distributed over it, alternatively having charge $\lambda = Q / \pi R^2$ per unit surface area.
We can calculate the strength of the $E$ field at a height $z$ above the center of the disc in the upwards direction as $$E(z) = \frac{\lambda}{2 \epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$$ or alternatively in terms of $Q$ as $$E(z) = \frac{Q}{2 \pi \epsilon_0 R^2} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right).$$
If I reduce $R$ to zero, the disc becomes a single point with charge $Q$. So I would expect the field strength to be proportional to $z^2$ like a normal point charge. What I see instead is a divide by zero problem.
Why am I not able to reduce this to a point charge by taking the limit $R \to 0$?
 A: There's no "divide by $0$" problem, instead you obtain a $\frac{0}{0}$ indefinite limit. The easiest thing to do in these cases is to use Taylor expansions (although you can probably use De l'Hôpital rule or any other trick you're familiar with); for instance we can calculate:
\begin{equation}
R^{-2}\bigg{(}1-\frac{z}{\sqrt{z^2-R^2}}\bigg{)} = R^{-2}\bigg{(}1-\frac{1}{\sqrt{1-\big{(}\frac{R}{z}\big{)}^2}}\bigg{)}\approx R^{-2}\bigg{(}1-\frac{1}{1-\frac{R^2}{2z^2}}\bigg{)}\approx R^{-2}\bigg{(}1-\big{(}1+\frac{R^2}{2z^2}\big{)}\bigg{)}=\frac{1}{2z^2}
\end{equation}
where $\approx$ means that we're neglecting $\mathcal{O}(R)$ terms, i.e. terms which go to $0$ when $R$ goes to $0$ and are thus negligible in the $R\rightarrow 0$ limit. Inserting this in the second equation you've written, you immediately obtain the field generated along the $z$ direction by a point charge:
\begin{equation}
E(z)=\frac{Q}{4\pi\epsilon_0z^2}
\end{equation}
A: \begin{align}
E(z) &=\frac{Q}{2 \pi \epsilon_0 R^2} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right) \cr
&=\frac{Q}{2 \pi \epsilon_0 R^2} \left(1 - \frac{1}{\sqrt{1 +(R/z)^2}}\right)\cr
&=\frac{Q}{2 \pi \epsilon_0 R^2} \left(1 - \frac{1}{1 +\frac{1}{2}(R/z)^2}\right)\qquad(R\ll z)\cr
&=\frac{Q}{2 \pi \epsilon_0 R^2} \left(1 - (1 -\frac{1}{2}(R/z)^2)\right) \cr
&=\frac{Q}{4 \pi \epsilon_0 z^2}
\end{align}
