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I want to know whether the function $\dfrac{x^2y}{x-y}$ is continuous at $(0,0)$ or not. when I switch to polar coordinate i get the limit of : $\dfrac{(r \cos^2 \theta \sin \theta)}{\cos \theta - \sin \theta}$ when r is going to zero. can I say that this limit does not exist because there are points where $\theta = \dfrac{\pi}{4}$ and then the denominator is zero, and therefore the function is not continuous at $(0,0)$ ? the value of the function when x is equal to y is 0 thank you

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    $\begingroup$ First notice that on the line $y = x$, i.e. $\theta = \dfrac{\pi}{4}$, this function is not defined. Now, if you take any path approaching to $\left( 0, 0 \right)$, i.e., $r \to 0$ and any $\theta$, then the limit is zero so that the limit at the function exists at $\left( 0, 0 \right)$. To know whether the function is continuous, we will also need the value of the function at $\left( 0, 0 \right)$. $\endgroup$ – Aniruddha Deshmukh Jun 22 at 9:29
  • $\begingroup$ I edit the post.. thank you $\endgroup$ – david Jun 22 at 9:38
  • $\begingroup$ With this, you can finally conclude that the function is continuous at $\left( 0, 0 \right)$. $\endgroup$ – Aniruddha Deshmukh Jun 22 at 9:39
  • $\begingroup$ @AniruddhaDeshmukh: It's not true that the limit is zero! See Josés answer below. $\endgroup$ – Hans Lundmark Jun 22 at 10:15
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Since$$\lim_{n\to\infty}f\left(\frac1n,\frac1n+\frac1{n^4}\right)=\lim_{n\to\infty}-\frac1{n^2}-n=-\infty,$$then, no matter how you define $f(0,0)$, the function $f$ will always be discontinuous at $(0,0)$.

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    $\begingroup$ Nice one. However, it should be pointed out that continuity is defined in the domain of existence of the function that is being considered. The poster asked about the continuity at $(0, 0)$ but actually the question of the continuity at that point does not arise, because the function does not exist there. Then, one may wonder about the possibilities to extend the function to a continuous one, and you have correctly ruled those out. $\endgroup$ – rubik Jun 22 at 9:58
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Let set $\ u=x-y\ $ then $$f(x,y)=\dfrac{x^2y}{x-y}=\dfrac{x^2(x-u)}{u}=\dfrac{x^3}u-\underbrace{x^2}_{\to 0}$$

So the behaviour in $(0,0)$ is determined by the relative "force" of $x$ and $u$.

This decomposition helps you finding counter examples.


For instance if $u\sim x^a$ then $\dfrac{x^3}{u}\sim x^{3-a}$

For $a>3$ this quantity tends to infinity, and we cannot extend by continuity.

  • this is J.C.Santos example above with $a=4$, let's take $x=\frac 1n$ then $y=\frac 1n-\frac 1{n^4}$

For $a=3$, you get a constant term, and this time we cannot extend by continuity because we can exhibit multiple limits.

  • indeed $f(\frac 1n,\frac 1n-\frac 1{\ell\ n^3})\to \ell\quad$ for any $\ell$ we want.
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