Does the limit exist? (Calculus) Consider the function 

$$f(x,y)=\frac{2xy^2\sin^2(y)}{(x^2+y^2)^2}.$$ 

Does the limit exist when $(x,y)$ tends to $(0,0)$?
 A: Let $x=\rho \cos \phi$ and $y=\rho \sin \phi$.  Then $$f(x,y)=\frac{2\rho^3\cos \phi \sin^2 \phi \sin^2(\rho \sin \phi)}{\rho^4}$$
Note that $\sin x \le x$.  Can you finish from here?
A: Just see this

$$ \frac{2xy^2 \sin^2(y)}{(x^2+y^2)^2}  \sim \frac{2 x y^4}{(x^2+y^2)^2} $$

since $\sin t \sim_{t\sim 0} t$. Then we have
$$ \bigg|  \frac{2 x y^4}{(x^2+y^2)^2}  \bigg| \leq   \frac{2 |x| |y|^4}{(x^2+y^2)^2} \leq \frac{2\sqrt{x^2+y^2}  (\sqrt{x^2+y^2})^4}{(x^2+y^2)^2} = 2\sqrt{x^2+y^2} < \epsilon $$
$$ \implies \sqrt{x^2+y^2} < \frac{\epsilon}{2} =\delta. $$
A: I also suggest the following approach. For example, if $x=\rho \cos \theta$ and $y=\rho \sin \theta$, the limit is zero. Then:
$$\left|\frac{2xy^2\sin^2(y)}{(x^2+y^2)^2}\right|\leq \left|\frac{2xy^2\sin^2(y)}{y^4}\right|= 2|x|\left|\frac{\sin^2(y)}{y^2}\right|\leq 2 |x|$$
Since $f(x,y)$ is between $0$ and $2 |x|$, the function is arbitrarily close to zero, when the distance between $(x,y)$ and $(0,0)$ is sufficiently small. It follows that the limit of $f$ is $0$, just to limit definition.
