WolframAlpha and I don't agree on $( xy\sin y )/(3x^2+y^2)$ as $(x,y)\to(0,0)$

Question

WolframAlpha claims that the following limit does not exist:

$$ \lim_{(x,y)\to(0,0)} \frac{xy\sin{y}}{3x^2+y^2} $$

https://www.wolframalpha.com/input/?i=lim+(xysiny)%2F(3x%5E2%2By%5E2)+as+%7Bx-%3E0,+y-%3E0%7D

However, I just proved that its limit is zero. Here is how:

1. $2|(\sqrt{3} x)(y)| \leq 3x^2 + y^2 $ by AM-GM.

2. Therefore, $ |xy \sin{y}| \leq \frac{1}{2\sqrt3}(3x^2+y^2)|\sin{y}|$

3. And so, $ 0 \leq |\frac{xy\sin{y}}{3x^2+y^2}| \leq \frac{1}{2\sqrt{3}}|\sin{y}|$

4. The one on the left and the right both go to zero as $(x,y)\to(0,0)$; therefore our expression in the middle must also converge to zero.
5. Finally, because $|\frac{xy\sin{y}}{3x^2+y^2}|$ converges to zero, $\frac{xy\sin{y}}{3x^2+y^2}$ converges to zero as well.

I am absolutely sure my proof is logical and correct, especially because WolframAlpha says '0' to this: https://www.wolframalpha.com/input/?i=lim+(xy%5E2)%2F(3x%5E2%2By%5E2)+as+%7Bx-%3E0,+y-%3E0%7D .

However, my textbook also claims it to be nonexistent.

What is the limit?

If it really is zero, then why does WolframAlpha claim it to be nonexistent?

If it isn't zero, where did I go wrong?

Answer 1

Yes, the limit is $0$. You can also prove that using the fact that$$\frac{xy\sin y}{3x^2+y^2}=\frac{xy^2}{3x^2+y^2}\times\frac{\sin y}y,$$that$$\lim_{(x,y)\to(0,0)}\frac{xy^2}{3x^2+y^2}=0$$ and that $\lim_{y\to0}\frac{\sin y}y=1$.

Answer 2

Yes we have that

$$\frac{xy\sin{y}}{3x^2+y^2}=\frac{\sin y}y\frac{xy^2}{3x^2+y^2} \to 1 \cdot 0=0$$

indeed by $x=\frac r {\sqrt 3} \cos \theta$ and $y=r \sin \theta$ we can easily show that $\frac{xy^2}{3x^2+y^2} \to 0$.

Answer 3

You are right.

To see the given limit exists and is zero, one may just employ polar coordinates, obtaining $$ \left|\frac{xy\sin{y}}{3x^2+y^2} \right|=\left|\frac{xy\sin{y}}{\color{red}{2x^2}+x^2+y^2} \right|\le \left|\frac{xy\sin{y}}{x^2+y^2} \right|= \frac{r^2\left|\cos \theta \sin \theta \right|\left|\sin{\left(r \sin \theta\right)} \right|}{r^2}\le \left|r \sin \theta\right|\le r $$ where we have made used of $$ \left|\sin \theta \right|\le1, \qquad\left|\cos \theta \sin \theta \right|\le1, \qquad|\sin x |\le |x|, \qquad x \to 0. $$