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I was doing exercises on finding if a multiple variable function is differentiable on a certain point (in this case, the origin) using the following limit:

$$ \lim_{\vec{v}\rightarrow\vec{0}}{\frac{f(\vec{a}+\vec{v})-f(\vec{a})-\vec{v}\cdot\vec{\nabla}f(\vec{a})}{||\vec{v}||}} $$

Where $\vec{v}$ is the difference vector and $\vec{a} = \vec{0}$. Knowing that if this limit tends to zero, $f$ is differentiable.

Well, the problem I found is that when searching for the gradient, the partial derivative respect to $x$ (on the origin) turned out to be $\infty$. My question is, what does that imply? Is then $f$ not differentiable?

This is the example where I had the problem:

$$ \begin{cases} f(x,y) = \frac{x^2 - y^2 + 2x^3}{x^2 + y^2}\;\text{, when}\;(x,y)\neq(0,0)\\ 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{, otherwise} \end{cases} $$ Doing the derivative we find:

$$ \frac{\partial f}{\partial x}=\lim_{h\rightarrow 0}{\frac{f(h,0)-f(0,0)}{h}}=\lim_{h\rightarrow 0}{\frac{h^2 + 2h^3}{h^3}}\rightarrow\infty $$

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1 Answer 1

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Change of coordinate system might help.

Use cylindrical coordinates, then

$\frac{x^2-y^2+2x^3}{x^2+y^2}=\cos{2\theta}+2r\cos^3{\theta}$

The $\theta$ dependence a the origin ($r=0$) implies the limit doesn't exist at the origin. So it's not continuous and so not differentiable.

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  • $\begingroup$ I think this proof doesn't work because of the dependence also on r. Moreover, you can easily prove with epsilon-delta that the function is continuous. $\endgroup$
    – D. Sarrat
    Nov 10, 2021 at 22:03
  • $\begingroup$ @D.Sarrat As an alternative approach, let $y=mx$. The limit of the resulting expression as x approaches 0 is $\frac{1-m^2}{1+m^2}$. Since the value depends on the direction from which you approach the origin, you don't have a limit. It looks like the limit is 0 as you approach from (1,1) and 1 if you approach from (1,0). $\endgroup$ Nov 10, 2021 at 22:12
  • $\begingroup$ But is the same problem. The limit not only depends on the parameter m, but it depends on x, so this doesn't apply. It's the same with cylindrical coordinates. $\endgroup$
    – D. Sarrat
    Nov 10, 2021 at 22:26
  • $\begingroup$ Ok never mind I've now understood it. Thanks ^^ $\endgroup$
    – D. Sarrat
    Nov 10, 2021 at 22:31
  • $\begingroup$ Suppose y is fixed at 0. What is the limit of this modified expression as x approaches zero? I get 1 valid with a 1D delta-epsilon proof. Let x be fixed at zero and let y approach 0. I get -1, consistent with the corresponding delta-epsilon argument. This implies path dependence on the limit. Similar to the issue here: math.stackexchange.com/questions/948563/… $\endgroup$ Nov 10, 2021 at 22:32

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