Electric field charged disc and L'Hôpital's rule I have been looking at the electric field of a charged disk and have a question about the use of l'Hopital's rule for the limiting case of electric field at points along the axis $z\gg$ disc radius $R$.  
$$E = \frac {q}{2\pi\epsilon R^2} \left(1 - \frac {z}{\sqrt{z^2+R^2}}\right)$$
I have applied l'Hôpital's rule in the limit of $R$ approaching zero, and see that the electric field approaches that of a point charge, as intuition suggests.  HOWEVER, when I use l'Hôpital's rule in the limit that $z$ approaches infinity, I get a repeating loop of indeterminate forms that doesn't arrive at the point charge expression.
My question is does this difference in results using l'Hopital's rule have any physical or mathematical significance?  
 A: To use L'Hôpital you either have to solve a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ kind of limit. I'll rewrite your expression to better show if this is the case:
$$E = \frac{q}{2 \pi \epsilon}\frac{\sqrt{z^2+R^2}-z}{R^2 \sqrt{z^2+R^2}}$$
(I just calculated the common denominator and separated the constants from the variables.)
As $R$ approaches 0, we can see that both numerator and denominator go toward 0, so we can use L'Hôpital.
Conversely, as z approaches infinity, we find $\frac{\infty - \infty}{\infty}$, so we cannot use L'Hôpital in this case. We first have to solve the $\infty - \infty$ indeterminate form. The easiest way (or the standard trick, if you prefer) is to multiply numerator and denominator by $\sqrt{z^2+R^2}+z$, so that the formula becomes:
$$E = \frac{q}{2 \pi \epsilon}\frac{R^2}{R^2 \sqrt{z^2+R^2}(\sqrt{z^2+R^2}+z)}$$
At this point $z$ disappears from the numerator and thus the indeterminate form is no more indeterminate and we can easily say that the limit goes to 0, as does the field of a point charge.
A: Update as the result of a comment from @garyp this time using L'Hôpital's rule.
$$E = \frac{q}{2 \pi \epsilon}\frac{(z^2+R^2)^{\frac 12}-z}{R^2 (z^2+R^2)^{\frac 12}}$$
Now differentiate twice with respect to $R$ the numerator and the denominator individually to get something like
$$ \frac {(z^2+R^2)^{-\frac 12} + R(.........)}{2(z^2+R^2)^{\frac 12} +R(.........)} $$
which has the limit $\dfrac{1}{2z^2}$ as $R$ tends to zero and gives the desired equation for the electric field due to a point charge.


$\left(1 - \dfrac {z}{\sqrt{z^2+R^2}}\right)$

Divide top and bottom of the fraction by $z$ and expand using the binomial theorem as far as the second term and for the electric field you will find that the $R^2$ cancels leaving the point charge field in terms of $z$.
