I am trying to solve whether $f(x,y) = \frac{xy}{(x^2+y)}$ is continuous at$(0,0)$ or not where $f(0,0)$ is defined to be $0$. I converted this to polar by substituting $x=r\cos p, y=r\sin p$ where $r$ tends to $0$ and got the result it is continuous. Book says it is not continuous. Where am I wrong and how to solve it?
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2How did you get the result it is continuous? You ask people to show where you are wrong without showing your work. – Jul 30 '16 at 19:43
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A good first start for these sorts of problems is to approach $(0,0)$ along a ray, say $t \mapsto (t, \alpha t)$. This will produce the limit $\alpha$ which clearly depends on how you approach $(0,0)$. – copper.hat Jul 30 '16 at 19:53
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You can use the polar method to show that it is NOT continuous. You CANNOT use the polar method to show it is continuous. – imranfat Jul 30 '16 at 20:12
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Let $x=t$ and $y=-t^2+h(t)$, where $h(t) \to 0$ as $t\to 0$, define a parametric curve. Then, we see that on this curve
$$\begin{align} f(x,y)&=\frac{xy}{x^2+y}\\\\ &=\frac{t(-t^2+h(t))}{t^2+(-t^2+h(t))}\\\\ &=\frac{t}{h(t)}(h(t)-t^2)\\\\ \end{align}$$
Now, if $h(t)=t^4$, for example, then we see that along this parametric curve
$$\begin{align} \lim_{(x,y)\to (0,0)}f(x,y)&=\lim_{t\to 0}f(t,-t^2+t^4)\\\\ &=\lim_{t\to 0^{\pm}}\left(t-t^{-1}\right)\\\\ &=\mp \infty \end{align}$$
and therefore, the function $f(x,y)$ cannot be continuous at $(0,0)$.
Mark Viola
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@donantonio Thank you. Much appreciative! But there is an easier way to show failure of existence. – Mark Viola Jul 30 '16 at 20:14