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

Question

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).

Answer 1

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

Answer 2

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

Answer 3

First of all, for all $y \neq 4$ you can write: $$\frac{xy-4x}{y^2-16} = \frac{x(y-4)}{(y-4)(y+4)} = \frac{x}{y+4},$$ and since limits deal with values of the function close to the point $(0,4)$ but not on $(0,4)$ exactly, we're good to go: $$\lim_{(x,y)\to(0,4)}\frac{xy-4x}{y^2-16} = \lim_{(x,y)\to(0,4)}\frac{x}{y+4} = 0.$$