Is there a discrete version of de l'Hôpital's rule? When considering asymptotics of runtime functions, you often have to find limits of quotients of discrete functions, e.g.
$\displaystyle\qquad \lim\limits_{n \to \infty} \frac{4^n}{\binom{2n}{n}\sqrt{n}}.$
While this particular case can easily be dealt with by Stirling's formula, I have been wondering. Mathematicians often like to use de l'Hôpital's rule, but it can obviously not be applied to the discrete case immediately (no mean value theorem). If—as in this case—you are lucky, you might find nice and well-studied continuations on the reals.
What to do in general, though? Is there a discrete version/relative of de l'Hôpital's rule, maybe using difference quotients?
 A: The discrete version of L'Hôspital's rule, in my opinion, is Abelian theorems, including the L'Hôspital's rule, Silverman-Toeplitz theorem and its sepcial case, Stolz-Cesàro theorem.
On de Bruijn's Asymptotic methods in analysis, it's said that

A theorem which derives asymptotic information about some kind of average of a function from asymptotic information about the function itself, is called an Abelian theorem. If one can find a supplementary condition under which the converse of an Abelian theorem holds, then this condition is called a Tauberian condition, and the converse theorem is called a Tauberian theorem.

The amount you gave, as far as I've tried, couldn't be easily solved with these theorems.
Let $$a_n=\frac{4^n}{\binom{2n}n\sqrt n}$$
we have
$$\ln a_{n+1}-\ln a_n=\frac12\ln(n+1)-\ln(n+\frac12)+\frac12\ln n=-\frac1{4n^2}+O\left(\frac1{n^3}\right)\tag1$$
Therefore $\ln a_n$ converges as $n\to\infty$. However, the preceding equation, the asymptotic behavior of the difference, is not enough to determine the limit value, even if the result is refined. Such efforts are generally unsuccessful.
However, if $S=\lim_{n\to\infty}\ln a_n$, we could determine the asymptotic behavior of $a_n-S$ through (1) easily, since $\ln a_n=S-\sum_{k\ge n}(\ln a_k-\ln a_{k+1})$.
Remark
One could determine $S$ through Stirling's formula. There's another approach, more elementary, I think:
$$a_n^2=\left(\frac{(2n-1)!!}{(2n)!!}\right)^2(2n+1)\cdot\frac n{2n+1}\to\frac1\pi$$
by Wallis product, therefore $S=1/\sqrt\pi$.
A: Stolz–Cesàro seems to be what you're looking for. There are two forms:
1.

Let $a_n$ and $b_n$ be two sequences approaching $0$ as $n\to\infty$, with $b_n$ decreasing. Then,
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
if the second limit exists.

2.

Let $a_n$ and $b_n$ be two sequences, with $b_n$ unbounded and increasing. Then,
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
if the second limit exists.

