Dividing multivariable Dirac delta function by single variable delta

Question

How can I divide $\frac{\frac{1}{8}\delta (x-0,y-0) + \frac{1}{4}\delta (x-2,y-0) + \frac{3}{8}\delta(x-0, y-4) + \frac{1}{4}\delta (x-2,y-4)}{\frac{1}{2}\delta(x-0) + \frac{1}{2}\delta (x-2)}$?

Answer 1

In the numerator is written as: $\frac{1}{8}\delta(x)(\delta(y)+3\delta(y-4))+\frac{1}{4}\delta(x-2)(\delta(y)+\delta(y-4))$. You use $\delta(x,y)=\delta(x)\delta(y)$ and simplify.

Now you can use the relation: $\frac{\delta(x)}{\delta(x)+\delta(x-2)}$ is zero for $x=2$ (because you divide by $\infty$ at this point) and 1 for $x \neq 0$ due to L'Hospital rule.

$\frac{\delta(x-2)}{\delta(x)+\delta(x-2)}$ is zero on $x=0$ and 1 elsewhere.