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?
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).