How do you compute the limit with multiple variables without fail? $$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}$$
I don't know what to do doesn't it give always $0$?
Whether $x=0$ $x=y$ or $y=0$ or $x=y^2$ it always give $0$ since it goes to $(0,0)$

HINT
Note that $$-2 \leq \dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \leq 2$$
Hence, $$(x^3 - y^3)\dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \in \left[-2\vert(x^3-y^3) \vert, 2\vert(x^3-y^3)\vert \right]$$
