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Checking continuity of $f(x,y)$ at $(0,0)$: $$ f(x,y)=\begin{cases}\dfrac{x^3+y^3}{x-y}\ \ ,x\neq y\\0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,x=y \end{cases}$$

Using polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$,

$$\lim_{r\rightarrow 0}\dfrac{r^3(\cos^3\theta+\sin^3\theta)}{r(\cos\theta-\sin\theta)}=0$$

Hence,$$\lim_{(x,y)\rightarrow (0,0)}f(x,y)=f(0,0)=0$$

But its graph doesn't look like continuous near-complete $z$-axis.

Being very new to this I don't know what's going on here, please help.

enter image description here

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  • $\begingroup$ The graph doesn't look very clear to me. But one thing one should forget is that one will always be able to reconcile every mathematical construct with one's intuition. For example, the Dirichlet function in real analysis is continuous at the origin, yet the graph doesn't look like it at all. Thus, once you can prove that your function is continuous, forget however the graph may look like. That's what our definitions lead us to. :) $\endgroup$ – Allawonder Aug 19 at 14:13
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    $\begingroup$ Aren't you incorrectly assuming $\dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded? $\endgroup$ – ganeshie8 Aug 19 at 14:21
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Your mistake is in falsely assuming that $g(\theta) = \dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded.

It has a discontinuity at $\theta = \pi/4$ so it is not bounded.
Hence you cannot conclude $ \lim\limits_{r\to 0}f(r) g(\theta)=0$

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  • $\begingroup$ I didn't knew that it needs to be bounded, my book didn't said that, anyway thanks for your answer, which books should I use to know more of these concepts? $\endgroup$ – mnulb Aug 19 at 14:38
  • $\begingroup$ @mnulb please see youtube.com/watch?v=9KrSU9E5PhY $\endgroup$ – ganeshie8 Aug 19 at 14:44
  • $\begingroup$ I'm not in academic... may not be the right person to suggest books :) sry.. But for this particular problem studying indeterminate forms is sufficient. Need to know $0\times \infty \ne 0$ $\endgroup$ – ganeshie8 Aug 19 at 14:45
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    $\begingroup$ @ganeshie8 Your last sentence is vague. Sometimes indeed $0×\infty$ may vanish. $\endgroup$ – Allawonder Aug 19 at 15:00
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    $\begingroup$ @Allawonder thanks haha should have been $0\times \infty $ may or may not be $0$ :) $\endgroup$ – ganeshie8 Aug 19 at 15:37
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Notice that taking the limit as $r \to 0$, like you did, only gets near zero through paths that are straight lines. This is not enough to ensure differentiability and this example is to show why. As pointed out in the comments, we have that $$ \frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta - \sin \theta} $$ is not bounded, when you approach zero through the line $y=x$. Although the function is defined to be zero on this line, you can still replicate this exploding behavior by choosing some path that approaches line $y=x$ sufficiently fast.

If you consider the path $x(t) = t$ and $y(t) = t - t^3$, then you have $$ \lim_{t \to 0} \frac{x^3 + y^3}{x - y} = \lim_{t \to 0} \frac{t^3 + (t - t^3)^3}{t^3} = \lim_{t \to 0}\frac{2t^3 - 3t^5 + 3t^7 - t^9 }{t^3} = 2. $$

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