Evaluate $\lim _{(x,y)\to(2,1)}\frac{\arcsin(xy-2)}{\arctan(3xy-6)}$ 
$$\lim _{(x,y)\to(2,1)}\frac{\arcsin(xy-2)}{\arctan(3xy-6)}$$

$$\lim _{(x,y)\to(2,1)}\frac{\arcsin(xy-2)}{\arctan(3xy-6)}\overset{z = xy}{=} \lim _{z\to 2}\frac{\arcsin(z-2)}{\arctan(3z-6)}=\lim _{z\to 2}\frac{\frac{1}{\sqrt{1-(z-2)}}}{\frac{3}{1+(3z-6)^2}}=\lim _{z\to 2}\frac{1+(3z-6)^2}{3\sqrt{1-(z-2)}}=\frac{1}{3}$$
Is the above answer valid?
 A: Yes it is a valid answer, by using L'Hospital's rule, just a typo at the end, it 's rather
$$
\lim _{z\to 2}\frac{\arcsin(z-2)}{\arctan(3z-6)}=\lim _{z\to 2}\frac{\frac{1}{\sqrt{1-(z-2)^{\color{red}{2}}}}}{\frac{3}{1+(3z-6)^2}}=\lim _{z\to 2}\frac{1+(3z-6)^2}{3\sqrt{1-(z-2)^{\color{red}{2}}}}=\frac{1}{3}.
$$
A: Yes, this is a valid approach.  For a slightly more rigorous justification, define:
$$
f(z) = \frac{\arcsin(z - 2)}{\arctan(3z - 6)}, \qquad g(x,y) = xy
$$
Because $g$ is a continuous function, we may conclude that
$$
\lim_{(x,y) \to (1,2)} f(g(x,y)) = \lim_{z \to g(1,2)} f(z)
$$
hence the validity of your result.
A: As everyone wrote, it is valid. For the limit itself, you could even avoid using L'Hospital rule writing $$\frac{\arcsin(z-2)}{\arctan(3(z-2))}=\frac{\arcsin(z-2)}{(z-2)}\times\frac{3(z-2)}{\arctan(3(z-2))}\times \frac13$$ You also could use the ratio of well known Taylor series $$\frac{\arcsin(z-2)}{\arctan(3(z-2))}=\frac{\arcsin(t)}{\arctan(3t)}=\frac{1}{3}+\frac{19 }{18}t^2+O\left(t^4\right)$$
