Q:

Dividing by epsilon and taking the limit of epsilon to zero

In deriving the joint distribution of two order statistics, there is the following step (F(x) is the Cumulative Dist Function at x, f(x) is the PDF at x):
$$
F(x-\epsilon)^{n-s}\times[F(x+\epsilon)-F(x-\epsilon)]\times [F(y-\epsilon)-F(x+\epsilon)]^{s-r-1} \times[F(y+\epsilon)-F(y-\epsilon)]\times (1-F(y+\epsilon))^{r-1}
$$
then, we divide by $\epsilon$ and take $\epsilon \to 0$, which yields:
$$
F(x)^{n-s} \times f(x)\times [F(y)-F(x)]^{s-r-1} \times f(y) \times (1-F(y))^{r-1}
$$
since 
$$
\lim_{\epsilon \to 0} \frac{F(x+\epsilon)-F(x-\epsilon)}{\epsilon} = f(x)
$$
My question is, there are two expressions using this epsilon to get $f(x)$ and $f(y)$, which means that its more like dividing by $\epsilon^2$?
I am confused on this point, is it that we can ignore this since $\epsilon$ is going to zero anyway? 

A:

You're right, this is wrong; the second expression is essentially the limit of the first expression divided by $\epsilon^2$ (up to two factors of $2$, as Henry pointed out).

