Find $\lim_{(x,y) \to (0,0)} \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$ 
Find $$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$$

What do you guys think about this approach, and are there any faster and easier approaches?
My approach:


*

*$\frac{\tan(x^3+y^3)(x^2+y^2)}{\sin(x^2+y^2)(x^2+y^2)}$
2.$\frac{\tan(x^3+y^3)}{(x^2+y^2)}$


*$\frac{\sin(x^3+y^3)}{\cos(x^3+y^3)\rightarrow 1(x^2+y^2)}$


*$\frac{\sin(x^3+y^3)}{(x^2+y^2)}$


*Polar: $rcos^3\phi + rsin^3\phi = 0 + 0 = 0$

Thus,
$$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$

Note that
$$ \frac{\tan(x^3 + y^3)}{\sin(x^2 + y^2)} = \frac{1}{\cos(x^3+y^3)} \frac{\sin(x^3 + y^3)}{\sin(x^2 + y^2)} = \underbrace{\frac{1}{\cos(x^3+y^3)}}_{\text{(1)}}
\underbrace{\frac{\sin(x^3 + y^3)}{x^3 + y^3}}_{\text{(2)}}
\underbrace{\frac{x^2 + y^2}{\sin(x^2 + y^2)}}_{\text{(3)}}
\underbrace{\frac{x^3 + y^3}{x^2 + y^2}}_{\text{(4)}}.$$
The first term in the product is continuous at $(x,y) = (0,0)$. The second term is of the form $g(p(x,y))$ where
$$ g(z) = \begin{cases} \frac{\sin(z)}{z} & z \neq 0 \\ 1 & z = 0 \end{cases} $$
is continuous and $p(x,y) \rightarrow 0$ as $(x,y) \rightarrow (0,0)$ and so tends to $g(0)$. The third term is also of the form $g(p(x,y))$ for a different $g$ and $p$. The last term can be dealt easily using polar coordinates.
