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 $(0,0)$
 A: HINT
Note that $$-2 \leq \dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \leq 2$$
Hence, $$(x^3 - y^3)\dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \in \left[-2\vert(x^3-y^3) \vert, 2\vert(x^3-y^3)\vert \right]$$
A: $\left |  \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right |\leq \dfrac{|x^3-y^3| (2x^2+3y^2)}{x^2+2y^2} \leq  \dfrac{|x^3-y^3| (2x^2+4y^2)}{x^2+2y^2}\leq 2|x^3-y^3|$
Where we used the inequality $|\sin (x)| \leq |x|$
ADDED Let $f(x,y)= x^3-y^3$, now knowing that $ \lim _{ (x,y) \rightarrow (0,0)} f(x,y)= 0$
It means that for $\epsilon >0$ there is a $\delta >0$ such that $ \forall (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$ we have $ | x^3-y^3| < \frac{ \epsilon }{2}$.
Therefore for that $\delta >0$  we have $\left |\dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right | < \epsilon$, whenever $ (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$. Since $\epsilon $ was arbitary we conclude be definition of the limit that 
$ \lim _{ (x,y) \rightarrow (0,0)}  \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}=0$. 
