limited limit 2

$\lim_{(x,y)\to(0,0)}\frac{3x-6y}{\sin(x-2y)} = \lim_{(x,y)\to(0,0)}\frac{3}{\frac{\sin(x-2y)}{x-2y}} = \boxed{3}$

This problem can rigorously be solved by putting it into one variable and applying L'Hopital's rule:

$\lim_{(x, y) \to (0, 0)} \frac{3x-6y}{sin(x-2y)} = \lim_{x \to 0} \frac{-3x}{sin(-x)} = \lim_{x \to 0} \frac{-3}{-cos(-x)} = 3$