<s>[INST] 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?
 [/INST] 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 $x^2+y^2 \ge 2xy$.
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