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Given:

$$ f(x,y) = \frac{xy^3}{x^2 + y^6}$$

Is it differentiable at $(0,0)$ ?

I said no, as it is not even continuous by the path $x = y^3$

However, when we go to polar view that is: $x = r \cos(\theta) , y = r \sin(\theta)$

we get that $$\lim_{r \rightarrow 0^+} \frac{r^4 \cos(\theta) \sin^3 (\theta)}{r^2(\cos^2(\theta) + r^4 \sin^6(\theta))}$$ and thus:

$$\lim_{r \rightarrow 0^+} \frac{r^2 \cos(\theta) \sin^3 (\theta)}{(\cos^2(\theta) + r^4 \sin^6(\theta)}$$

And we can just plot $r = 0$ and get that it is indeed continuous...

What is wrong with this way? I don't understand, as we are taught to use this way to prove or disprove continuity every time, but I did not check if it actually worked or not! I assumed this way works every time, so why does it fail here?

Toby Mak
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CSch of x
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1 Answers1

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The problem is that : $\lim_{r \rightarrow 0^+} \frac{r^2 \cos(\theta) \sin^3 (\theta)}{(\cos^2(\theta) + r^4 \sin^4(\theta)}$ is not always determinate contrary to what you probably think. What happens if $\theta$ assumes a value which makes denominator $0$ as $r\to 0$. You have indeterminate form $(0/0)$.
As it turns out, particularly in this case, the above limit depends upon $\theta$, i.e., the direction through which you approach $(0,0)$. Try doing $r\to 0$ along $\theta =\pi/2$. See what happens.

Koro
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  • But from what I know, in some cases we use this way "blindly" without checking - this is just "luck" that I checked $x=y^3$ , how do you suppose to "catch" it before you hand in the test? – CSch of x Aug 24 '20 at 11:01
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    @StackOMeow:Just examine: Is the expression in polar coordinates free of $\theta$? If yes, then you can evaluate easily using $r\to $" wherever it tends to" but if not, then the value of limit may be dependent upon $\theta$ that is limit doesn't exist. – Koro Aug 24 '20 at 11:09