The limit of a two-variable function $\lim_{ (x, y) \to (0,0) } \frac{ xy \sin y }{ x^2 + y^2 }$ I'm trying to check if this two-variable function has a limit on the point $(0,0)$:
$$\lim _{ (x, y) \to (0,0) } {{ xy \sin y } \over { x^2 + y^2 }}$$
So my method was:
$$ {{xy \sin y} \over {x^2 + y^2}} = {{y \sin y} \over y^2} {x \over { ( {x^2 \over y^2 }) + 1 }} $$
Clearly
$$ \lim_{y \to 0} {y \sin y \over y^2} = 1 $$
But
$$ {x \over { ( {x^2 \over y^2 }) + 1 }} = { xy^2 \over x^2 + y^2 } $$
Moreover
$$ | { xy^2 \over x^2 + y^2 } | < |x| $$
and $\lim_{x \to 0} x = 0$.
Thus
$$ \lim _{(x, y) \to (0, 0)} {x \over { ( {x^2 \over y^2 }) + 1 }} = 0 $$
Hence
$$\lim _{ (x, y) \to (0,0) } {{ xy \sin y } \over { x^2 + y^2 }} = 1 \times 0 = 0$$
But calculators say that the limit does not exist.
I wonder where I have done wrong.
 A: $$x = r\cos(\theta)$$
$$y = r\sin(\theta)$$
$$\lim _{ (x, y) \to (0,0) } {{ xy \sin y } \over { x^2 + y^2 }}=\lim _{r\to \:0}\left(\frac{r^2\cos \left(θ\right)\sin \left(θ\right)\sin \left(r\sin \left(θ\right)\right)}{\left(r\cos \left(θ\right)\right)^2+\left(r\sin \left(θ\right)\right)^2}\right)$$
$$ = \lim _{r\to \:0}\left(\cos \left(θ\right)\sin \left(θ\right)\sin \left(r\sin \left(θ\right)\right)\right)=\color{red}{0}$$
A: The limit does exist and your approach is fine. The problem is with your calculator. Calculators and math software in general are notoriously bad at finding multidimensional limits compared to normal one dimensional limits (it's just a very hard problem to determine such limits on a computer).
If I ask Wolfram|Alpha about the limit $\lim_{(x,y)\to (0,0)} \frac{xy}{x^2+y^2}\sin(y)$ it gives me "(limit does not exist, is path dependent, or cannot be determined)". If I change $\sin(y)$ to $y$ and ask what is the limit $\lim_{(x,y)\to (0,0)}\frac{xy}{x^2+y^2}y$ it gives me $0$ as the result. However these two limits have to be the same as $\frac{\sin(y)}{y} \to 1$.
It's anyway good to see that Wolfram|Alpha has added the "cannot be determined" part to the answer. A year ago it simply (and wrongly) would have said that the limit does not exist (see e.g. this related question).
