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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.$

plhn
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    what do you really have you in your mind when you post this just after asking http://math.stackexchange.com/questions/572098/show-continuity-of-fracxy-sqrtx2y2 ? –  Nov 18 '13 at 17:30
  • @PraphullaKoushik I'm just reading a math book. this is a consecutive question. um... by any chance, this can be solved by that inequality? you mean that? – plhn Nov 18 '13 at 17:34
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    you need not go in that approach (I am not sure even if you can use that).. what generally one use in this approach is, you just consider limit $(x,y)\rightarrow (0,0)$ in some two or three ways i mean two or three possible paths and then you are done... –  Nov 18 '13 at 17:38
  • @plhn: PraphullaKoushik is right. Check out my answer. – meta_warrior Nov 18 '13 at 17:40

3 Answers3

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HINT: Let $f$ approach $(0,0)$ along $y=x$, so $f=\frac{x^2}{2x^2}=\frac{1}{2}$ along $y=x$

meta_warrior
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$y=mx\Rightarrow f(x,y)=\frac{mx^2}{(1+m^2)x^2}=\frac{m}{1+m^2}$

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....

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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 $(0,0)$.

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