Is the function $f(x,y) = \begin{cases} \frac{x \sqrt{|y|} }{2x^2+|y|} &(x,y) \neq (0,0) \\0&(x,y) = (0,0)\end{cases}$ continuous at $R^2$? I have the following function
$$f(x,y) = \begin{cases} \frac{x \sqrt{|y|} }{2x^2+|y|} &(x,y) \neq (0,0) \\0&(x,y) = (0,0)\end{cases}$$
Is this function continuous at $R^2$?
I've tried to prove that it is not continuous at $(0,0)$, but did not succeed.
Would much appreciate help.
Thank you!
 A: The limit at the origin along the positive arm of the path $y=x^2$ is $\dfrac{1}{3}$. So it is not continuous there.
$$ \lim_{x\to0^+}\frac{x\sqrt{x^2}}{2x^2+x^2}=\lim_{x\to0^+}\frac{|x|}{3x}=\frac{1}{3}$$
A: For a function to be continuous at $(x_0,y_0)$ in $\mathbb{R}^2$ it is not enough that the partial limits in the axes $x=0$ and $y=0$ exist and be equal, as I'm supposing you tried (it's a common mistake). You need to guarantee that this limit is the same for all paths going through the point.
For instance, in this case, the axes limits are both equal:
$$
\lim\limits_{y\to0}f(0,y)=\lim\limits_{x\to0}f(x,0)=0.
$$
But you can also take the parabola $y=x^2$ and check that
$$
\lim\limits_{x\to0^\pm}f(x,x^2)=\lim\limits_{x->0^\pm}\frac{x|x|}{3x^2}=\pm\frac{1}{3}.
$$
Since they are different, the limit doesn't exist.
A little visualization goes a long way, I recommend Mathematica if you have access to a license.

A: We have that by $x=u$ and $|y|=v^2$
$$ \frac{x \sqrt{|y|} }{2x^2+|y|} = \frac{u v }{2u^2+v^2}=\frac{\cos \theta \sin \theta}{\cos^2\theta + 1}$$
therefore the limit doesn't exists at $(0,0)$.
