Find $\lim_{(x,y)\to (3,1)\\(x≠3y)} \frac{3x^2-5xy-5y^2+2x-4y}{x-3y}$ 
$$\lim_{(x,y)\to (3,1)\\(x≠3y)} \frac{3x^2-5xy-5y^2+2x-4y}{x-3y}$$

My work so far
$$\lim_{x\to 3} \frac{3x^2-5xy-5y^2+2x-4y}{x-3y}$$
$$\lim_{x\to 3} \frac{3(3)^2-5(3)y-5y^2+2(3)-4y}{3-3y}$$
$$=\frac{-5y^2-19y+33}{3-3y}$$
and
$$\lim_{y\to 1} \frac{3x^2-5xy-5y^2+2x-4y}{x-3y}$$
$$\lim_{y\to 1} \frac{3x^2-5x(1)-5(1)^2+2x-4(1)}{x-3(1)}$$
$$=\frac{3x^2-3x-9}{x-3}$$
Would anything else need to be done? Also, if my process correct?

the iterated limits do not exist, as it tends to infinity. so the limit does not exist, otherwise it would not matter how you approach the point it would tend to some point $(x_{0},y_{0})$ $\in$ $\mathbb{R}^2$.
