Show discontinuity of $\frac{xy}{x^2+y^2}$ How to show this function's discontinuity?
$ f(n) = \left\{ 
   \begin{array}{l l}
     \frac{xy}{x^2+y^2} & \quad , \quad(x,y)\neq(0,0)\\
     0 & \quad , \quad(x,y)=(0,0)
   \end{array} \right.$
 A: HINT: Let $f$ approach $(0,0)$ along $y=x$, so $f=\frac{x^2}{2x^2}=\frac{1}{2}$ along $y=x$
A: $y=mx\Rightarrow f(x,y)=\frac{mx^2}{(1+m^2)x^2}=\frac{m}{1+m^2}$
Does this ring some bell??? 
I am actually trying to make you more comfortable with this kind of idea.
In this case you know limit should be zero and if you consider path $y=x$ then you are done.
But, then more generally what would one do if you are not so sure of $\lim$ is you keep changing $m$ in $y=mx$ (some other curve in some other case) and see if the result is same even if i change $m$.
In this case we got $f(x,y)=\frac{m}{1+m^2}$.
If i approach through $y=x$ keeping $m=1$, I would get the result as $f(x,y)=\frac{1}{2}$
If i approach through $y=2x$ keeping $m=2$, I would get the result as $f(x,y)=\frac{2}{5}$
So, If i approach in two different paths those limits are not equal. So for get about $\lim _{(x,y)\rightarrow (0,0)}f(x,y) $ being equal to $f(0,0)$  (which is what you need for continuity), The limit $\lim _{(x,y)\rightarrow (0,0)}f(x,y) $ does not even exist.
So, function is discontinuous.... 
A: As in here, use polar coordinates. Your limit is equivalent to
$$\lim_{\rho\rightarrow 0}\frac{\rho^2\cos\theta\sin\theta}{\rho^2}=
\cos\theta\sin\theta, $$
whose value depends on the angle $\theta$. So the function is not continuous at $(0,0)$.
