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 $(0,0)$?

Question

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.

Answer 1

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

$$\sin(y)+\frac{y^2}{x^2+y^2}(\sin(x)-\sin(y))$$

and, by MVT, $$|\sin(x)-\sin(y)|\le |x-y|$$

thus $$|f(x,y)-0|\le |y|+|x-y|$$ $$\le |x|+2|y|$$

So, $$0<\sqrt{x^2+y^2}<\delta\implies$$ $$|x|<\delta \text{ and } |y|<\delta\implies$$ $$|f(x,y)-0|<3\delta$$

from here, we see that we can take $$\delta=\frac{\epsilon}{3}$$

Answer 2

It is enough to show that the limit of $\frac{x^2 \sin y}{x^2+y^2}$ is $0$ ( the other half follows by symmetry). The modulus of this is at most $|\sin y|$, which is less than $|y|$, so we are done.

Answer 3

As suggested in the comments, by polar coordinates

$$\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}=\cos^2\theta \sin (r \sin \theta)+\sin^2\theta \sin (r \cos \theta) \to 0$$

Answer 4

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 $\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}\sin y+y^{2}\sin x}{x^{2}+y^{2}}=0$.

Answer 5

If the limit exists, it will exist for any metric (so long as it is a valid metric).

While it may seem natural to use the Euclidean metric, it is not always the easiest to work with. I frequently prefer to work with the maximum metric.

That is $d((x_1,y_1),(x_2,y_2)) = \max (|x_2-x_1|,|y_2-y_1|)$

In this example, $\delta = \max (|x|,|y|)$

and $\left|\frac {x^2\sin y + y^2\sin x}{x^2 + y^2}\right| < \frac {2\delta^3}{\delta^2} = 2\delta$