# How do I prove differentiability of a multiple variable function when the partial derivatives tend to infinity?

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:

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

## 1 Answer

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.

• 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. Nov 10, 2021 at 22:03
• @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). Nov 10, 2021 at 22:12
• 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. Nov 10, 2021 at 22:26
• Ok never mind I've now understood it. Thanks ^^ Nov 10, 2021 at 22:31
• 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/… Nov 10, 2021 at 22:32