Q:

Calculating $\lim_{(x,y) \to (0,4)} \frac{xy-4x}{y^2-16}$ or proving it does not exist

I've managed to get the limit into the following form:
$y=mx \rightarrow \lim_{x \to 0} \frac{mx^2-4x}{mx^2-16} \rightarrow \lim_{x \to 0} \frac{x (m-4)}{x^2 (m+4)  (m-4)}$
I'm not sure how I'm supposed to procceed from here (and might be on the wrong path altogether).

A:

We have $$\lim_{(x,y) \to (0,4)} \frac{xy-4x}{y^2-16}$$ $$=\lim_{(x,y) \to (0,4)} \frac{x(y-4)}{(y-4)(y+4)}$$ $$=\lim_{(x,y) \to (0,4)} \frac{x}{(y+4)}$$ $$= \frac{0}{(4+4)}=0$$

A:

$$\frac{xy-4x}{y^2-16} = \frac{x(y-4)}{(y+4)(y-4)} = \frac{x}{y+4}$$
So we now have $$\lim_{x,y \to (0,4)} \frac{xy-4x}{y^2-16} = \lim_{x,y \to (0,4)}\frac{x}{y+4} = \frac08 = \boxed{0}$$

