I need to use Taylor's theorem to compute: $\lim\limits_{(x,y)\to (0,0)}{\sin(xy)-xy\over x^2y}$ Using the theorem we have that: $\sin(xy)=xy-{(xy)^3\over6}+R_6(x,y)$ where $\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over (x^2+y^2)^3}=0$

So $$\lim\limits_{(x,y)\to (0,0)}{\sin(xy)-xy\over x^2y}=\lim\limits_{(x,y)\to (0,0)}-{{(xy)^3\over6}+R_6(x,y)\over x^2y} \\[2em] =\lim_{(x,y)\to (0,0)}-xy^2/6+\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over x^2y}$$

But my problem is that I don't know how to compute $\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over x^2y}$

Any hints, suggestions or ideas would be highly appreciated.