<s>[INST] How to prove that the limit of $f(x,y)=\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ is $0$ as $(x,y)$ approaches $(0,0)$? I need to show that $\displaystyle\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ exists. I know that it is equal to zero. Until now, all I know how to do is to prove using the $\epsilon-\delta$ definition. I tried to prove it by the following way:
Let $\delta>0$ be a real number such that $0<\sqrt{x^2+y^2}<\delta$. We have $0< x^2+y^2<\delta^2$.
We know that $\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq\left\vert x^2\sin(y)\right\vert+\left\vert y^2\sin(x)\right\vert=x^2\left\vert\sin(y)\right\vert+y^2\left\vert\sin(x)\right\vert\leq x^2+y^2$.
Thus
$$0<\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq x^2+y^2<\delta^2$$
and dividing the inequality by $x^2+y^2>0$ we have
$$ 0<\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}<1<\dfrac{\delta^2}{x^2+y^2}. $$
I thought that this could be useful because $\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}=\left\vert\dfrac{x^2\sin(y)+y^2\sin(x)}{x^2+y^2}\right\vert$, but I don't know how to proceed.
 [/INST] Recall that $|\sin t|\leq |t|$ for any $t\in\mathbb{R}$. For any $x,y\in\mathbb{R},$
we have that
\begin{eqnarray*}
 &  & \left|x^{2}\sin y+y^{2}\sin x\right|\\
 & \leq & \left|x^{2}\sin y\right|+\left|y^{2}\sin x\right|\\
 & \leq & |x^{2}y|+|y^{2}x|.
\end{eqnarray*}
Now, let $\varepsilon>0$ be given. Define $\delta=\frac{1}{2}\varepsilon$.
Let $(x,y)\in\mathbb{R}$ that satisfies $0<\sqrt{x^{2}+y^{2}}<\delta$.
We have estimation:
\begin{eqnarray*}
 &  & \left|\frac{x^{2}\sin y+y^{2}\sin x}{x^{2}+y^{2}}\right|\\
 & \leq & \left|\frac{x^{2}}{x^{2}+y^{2}}\right||y|+\left|\frac{y^{2}}{x^{2}+y^{2}}\right||x|\\
 & \leq & |y|+|x|\\
 & \leq & \sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+y^{2}}\\
 & < & 2\delta\\
 & = & \varepsilon.
\end{eqnarray*}
This shows that $\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}\sin y+y^{2}\sin x}{x^{2}+y^{2}}=0$.
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