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,
$$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$
 A: 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.
A: I would say for $\vert z \vert$ small enough $$\vert \tan z \vert \le 2 \vert z \vert$$
Hence for $\vert x \vert + \vert y \vert$ small enough
$$\left\vert \frac{\tan(x^3+ y^3)}{\sin (x^2 +y^2)} \right\vert \le \frac{2(\vert x \vert^3+ \vert y \vert^3)}{\sin (x^2 +y^2)} \le 2 \frac{x^2+y^2}{\sin (x^2+y^2)}(\vert x \vert + \vert y \vert)$$ and you are done as $\lim\limits_{z \to 0} \frac{\sin z}{z} = 1$ and $\lim\limits_{(x,y) \to 0} \vert x \vert + \vert y \vert =0$
