Assistance in showing the limit of $\lim_{(x,y) \to (0,0)} \frac{x \sin^{2}y}{x^{2} + y^{2}}$ exists. I've been asked to decide if the following limit exists and provide justification.
$$\lim_{(x,y) \to (0,0)} \frac{x \sin^{2}y}{x^{2} + y^{2}}$$
One idea that I had and it seems to be giving me the closest chance of success was to do the following:
$$\frac{x \sin^{2}y}{x^{2} + y^{2}}  \leq x\ \frac{\sin\ y}{y} \ \frac{\sin\ y}{y}$$
Now from single variable calculus we know $\lim_{y \to 0}\frac{\sin\ y}{y} = 1$ and clearly $\lim_{x \to 0} x = 0$. My problem though is trying to squeeze it from below. It is not necessarily the case that the original function is always positive because of that pesky single $x$ value in the numerator.
A hint we were given was to remember the behaviour of $|\sin\ y|$ that would control the function. Which I feel would be the idea that $|\sin\ y| \leq 1$. Using that idea I would still end up in a situation where I'm left with
$$\frac{x \sin^{2}y}{x^{2} + y^{2}}  \leq \frac{x}{x^2 + y^2}$$
and I'm not sure how I can control that to my benefit.
Could I get some sort of guiding hint or tip on how to approach this if I'm not using the right ideas?
 A: Let $\varepsilon>0$ be given. Define $\delta=\frac{\varepsilon}{2}.$
We go to show that $\left|\frac{x\sin^{2}y}{x^{2}+y^{2}}-0\right|<\varepsilon$
whenever $(x,y)\in B((0,0),\delta)\setminus\{(0,0)\}.$ Let $(x,y)\in B((0,0),\delta))\setminus\{(0,0)\}$
be arbitrary, i.e, $0<x^{2}+y^{2}<\delta^{2}$. Choose $r>0$ and
$\theta\in\mathbb{R}$ such that $x=r\cos\theta$ and $y=r\sin\theta$.
Note that $r=\sqrt{x^{2}+y^{2}}<\delta$. Since $|\sin y|\leq|y|$,
we have that
\begin{eqnarray*}
\left|\frac{x\sin^{2}y}{x^{2}+y^{2}}-0\right| & \leq & \left|\frac{xy^{2}}{x^{2}+y^{2}}\right|\\
 & = & \left|\frac{r^{3}\cos \theta\sin^{2}\theta}{r^{2}}\right|\\
 & \leq & r\\
 & < & \delta\\
 & = & \frac{\varepsilon}{2}\\
 & < & \varepsilon.
\end{eqnarray*}
Therefore, $\lim_{(x,y)\rightarrow(0,0)}\frac{x\sin^{2}y}{x^{2}+y^{2}}=0.$
A: By using the following inequality
$$
\biggl|\frac{x\sin^2y}{x^2+y^2}\biggr|\le \frac{|x|y^2}{x^2+y^2}
$$
and substituting $x=r\cos\theta$ and $y=r\sin \theta$, you can get to the expected result.
A: Hint: Notice that, $$\left|\dfrac{x\sin^2y}{x^2+y^2}\right|\leq \dfrac{|x|y^2}{x^2+y^2}.$$ We have that $(|x|-y)^2\geq 0$, so, $x^2-2|x|y+y^2\geq 0$, that is, $x^2+y^2\geq 2|x|y$. Hence, $$\dfrac{1}{x^2+y^2}\leq \dfrac{1}{2|x|y}.$$ Thus, $$\left|\dfrac{x\sin^2y}{x^2+y^2}\right|\leq \dfrac{y}{2}.$$
