Limit as $(x,y) \to (0,0)$ of $\frac{x^2\sin(x)}{x^2 + y^2}$

Question

I need to find $$\lim_{(x.y) \to (0,0)}\frac{x^2 \sin(x)}{x^2 + y^2}$$
Wolfram says that this limit is undefined. However, I attempted to solve this and I got that the limit is $0$. Therefore, I'd be grateful if you could tell me where my reasoning went wrong.

Since the limit is at the origin, I can apply polar coordinates:
$$\lim_{r \to 0} \frac{r^2 \sin^2(\theta) \sin(r \sin(\theta))}{r^2(\sin^2(\theta)+\cos^2(\theta))} = \lim_{r \to 0}\sin^2(\theta) \sin(r \sin(\theta))$$ Now, we know that this expression $\sin(\theta) $ is bounded. Therefore $r \sin(\theta)$ approaches $0$ since $r$ approaches $0$. This entails that $\sin(r \sin(\theta)) $ approaches $0$ because $\sin(0) = 0$. Finally. $\sin^2(\theta) $ is bounded, and so the limit in questoin becomes $0$.

Where is the error in my reasoning?

Answer 1

Observe that $\forall x\ne 0$ and $\forall y\in\Bbb R,$

$$x^2+y^2\ge x^2>0$$

$$\implies $$

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

thus the limit is zero.

Answer 2

$| { x^2 \sin x \over x^2+y^2}| \le| { x^3 \over x^2+y^2}| \le |x| | { x^2 \over x^2+y^2}| \le |x|$.