# Not sure that this function isn't differentiable at the origin

For a fixed $k\in\mathbb{N}$, define $f_k\colon\mathbb{R}^2\to\mathbb{R}$ by: $$f_k(x,y) = > \begin{cases}f_k(x,y)=\frac{x^2(x+y^2)}{x^2+y^{2k}}\mbox{ > if}~(x,y)\neq(0,0)\\0\mbox{ if} ~(x,y)=(0,0).\end{cases}$$ Show that $f_1$ is not differentiable at $(0,0)$, but $f_k$ is differentiable at $(0,0)$ for each $k\geq 2$.

I have started with $f_1 = \frac{x^3+x^2y^2}{x^2+y^2}$. $$\frac{\partial f_1}{\partial x}(0,0) = \lim\limits_{h\to0}\frac{f_1(0+h,0)-f_1(0,0)}{h} = \lim\limits_{h\to0}\frac{\frac{h^3}{h^2}-0}{h} = 1$$ $$\frac{\partial f_1}{\partial y}(0,0) = \lim\limits_{h\to0}\frac{f_1(0,0+h)-f_1(0,0)}{h} = 0$$

So here are both partial derivatives. Denote $\theta=(0,0)$. Now I need to check that $$\lim\limits_{h\to0}\frac{f_1(\theta+h)-f_1(\theta)-\bigtriangledown f1(\theta)\cdot h}{||h||} = 0$$ for differentiability at the origin.

$$\lim\limits_{h\to0}\frac{f_1(\theta+h)-f_1(\theta)-\bigtriangledown f1(\theta)\cdot h}{||h||} = \lim\limits_{h\to0}\frac{\frac{h^3}{h^2}-0-h}{||h||} = 0$$

Where I had made a mistake? It isn't supposed to be differentiable.

What is the best way to prove diffentiability of $f_k$ at the origin for arbitrary $k\geq 2$ further? Am I supposed to calculate partial derivatives of $f_k$ with respect to $x$ and $y$ and show that they are continuous?

Any help will be appreciated.

• $h$ is a 2-dimensional vector, $h=(h_1,h_2)$. Expressions like $h^3/h^2$ do not makes sense. – Andrés E. Caicedo Aug 14 '18 at 18:38

The mistake is that you wrote $f_1(\theta+h)$=$\frac{h^3}{h^2}$. When you calculate the partial derivative $\frac{\partial f_1}{\partial x}$ then you really take the limit when $y=0$. But when you try to prove differentiability $h$ is not a number, it's a vector. So here I'll recommend to work with $x$ and $y$ instead of $h=(h_1,h_2)$. You need to calculate the following limit:
$\lim\limits_{(x,y)\to(0,0)}\frac{f_1(x,y)-f_1(0,0)-\frac{\partial f_1}{\partial x}(0,0)*x-\frac{\partial f_1}{\partial y}(0,0)*y}{\sqrt{x^2+y^2}}$. You know all the expressions, so just calculate it.