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

I tried multiplying $x^3y^3$ the nominator and denominator, didn't work.
I tried the polar way too, $x = r \cos\phi$, $y=r \sin\phi$,$x^2 + y^2 = r^2$ too. which gives me $\frac{"0"}{"0"}$
I tried proving that this limit doesn't exist but wolfram shows that its limit is zero.
Can anyone give me a hint on how to approach this?
 A: Switching to polar is a good idea and gives
$$\lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4}$$
You probably know that $\sin(a)/a$ tends to 1 for $a \to 0$; so:
$$\begin{array}{cc}
\displaystyle \lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4} & \displaystyle
= \lim_{r \to 0} \left( \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^5 \cos^3t\sin^2t}(r \cos^3t\sin^2t) \right) \\ \\
& \displaystyle 
= \underbrace{\lim_{r \to 0}  \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^5 \cos^3t\sin^2t}}_{\to 1} \underbrace{\lim_{r \to 0}  \left( r \cos^3t\sin^2t \right)}_{\to \ldots} 
\end{array}$$
Can you take it from here?

Alternatively:
$$\lim_{(x,y)\to(0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}
=\lim_{(x,y)\to(0,0)} \frac{\sin(x^3y^2)}{x^3y^2}\frac{x^3y^2}{(x^2+y^2)^2}
=\lim_{(x,y)\to(0,0)} \frac{x^3y^2}{(x^2+y^2)^2} = \cdots$$
And use $x^2+y^2 \ge 2xy$.
A: You can use $|\sin t| \le |t|$ and AM-GM inequality:
$$
\left|\frac{\sin(x^3 y^2)}{(x^2+y^2)^2}\right|\le \left|\frac{x^3y^2}{(x^2+y^2)^2}\right|\le \left|\frac{x^3y^2}{(2xy)^2}\right|=\frac{1}{4}|x|
$$
and apply sandwich theorem.
A: $$\frac{\sin(x^3y^2)}{(x^2+y^2)^2}=\frac{x^3y^2}{(x^2+y^2)^2}\frac{\sin(x^3y^2)}{x^3y^2}$$
The second factor above tends to $\;1\;$ and as for the first one: with polar coordinates:
$$\frac{x^3y^2}{(x^2+y^2)^2}\rightarrow r\cos^3\theta\sin^2\theta\xrightarrow[r\to0]{}0$$
Thus the whole thing tends to zero
A: Hint
As I wrote in comments, move along $y=kx$. Then $$ A=\frac{\sin(x^3y^2)}{(x^2+y^2)^2}=\frac{\sin \left(k^2 x^5\right)}{\left(k^2+1\right)^2 x^4}$$ Now, for small $z$, $\sin(z]\approx z$ So $$A \approx \frac{k^2 x^5}{\left(k^2+1\right)^2 x^4}=\frac{k^2 x}{\left(k^2+1\right)^2 }$$
