Show that $f$ is not differentiable at $(0,0)$ - $\frac{x_1^2x_2}{x_1^2+x_2^2}$

Question

Let the function $f: \mathbb{R^2} \to \mathbb{R}$ defined as

$$f(x) = \begin{cases} \frac{x_1^2x_2}{x_1^2+x_2^2}, & \quad \text{if } (x_1,x_2) \not= 0 \\ 0, & \quad \text{if } (x_1,x_2) = 0 \\ \end{cases} $$

Show that $f$ is not differentiable at $(0,0)$.

I think I can use the directionnal derivative or the polar coordinate, but I don't know how.

Is anyone is able to explain to me how I can use these properties? If I can't, could you propose to me an alternative?

P.S. Please, don't try to use a very specific analysis theory; I am only an undergraduate student (bachelor).

Answer 1

To check whether a function $f: \Bbb{R^2}\to \Bbb{R}$ is differentiable at $(0,0)$, we need to find a linear map $Df(x_0)$, called differential. Now, let's take $u=(u_1,u_2)\in \Bbb{R^2-\{(0,0)\}}$. The directional derivatives along $u$ are:

$$D_uf(0,0)=\lim_{t\to 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(tu_1,tu_2)}{t}=\lim_{t\to 0}\frac{t^3u_1^2u_2}{t^3(u^2_1+u^2_2)}=\frac{u^2_1u_2}{u_1^2+u_2^2}$$

So directional derivatives exist, and in particular, for $u=(1,0)$ and $u=(0,1)$, partial derivatives exist at $(0,0)$ and are equal to $0$. So our main candidate for the differential is $Df(0,0)(u_1,u_2)=0$. Then, in order to prove the differentiability of $f$ at $(0,0)$, we must prove that

$$\lim_{(x_1,x_2)\to (0,0)}\frac{x_1^2x_2}{(x_1^2+x_2^2)\sqrt{x_1^2+x_2^2}}=0$$

We'll see that this does not happen. Let $x_1=r\cos\theta$, $x_2=r\sin\theta$, Then

$$\lim_{(x_1,x_2)\to (0,0)}\frac{x_1^2x_2}{(x_1^2+x_2^2)\sqrt{x_1^2+x_2^2}}=\lim_{r\to 0}\frac{r^3\cos^2\theta \sin\theta}{r^3}=\cos^2\theta \sin\theta$$

So the limit depends on $\theta$, and therefore, it doesn't exist, so $f$ is not differentiable at $(0,0)$.

Answer 2

Starting like A. Jiménez, $$D_uf(0,0) = \frac{u^2_1u_2}{u_1^2+u_2^2}.$$ But can be easily proved that if $f$ is differentiable in $(0,0)$: $$D_uf(0,0) = Df(0,0)u,$$ and this is impossible because in this case $D_uf(0,0)$ is a nonlinear funcion of $u$.