Evaluating partial differentiability, sequences in $\mathbb R^{2}$

Question

Let $f: \mathbb R^{2} \to \mathbb R$

$ f(x_{1},x_{2})$=

\begin{cases} 0 & x= 0 \\ \frac{x_{1}x_{2}^{2}}{\vert\vert x\vert\vert_{2}^{4}} & x \neq 0 \\ \end{cases}

Show:

(a) $f$ is continuous in $x\in\mathbb R^{2}-\{0\}$.

(b) $\forall$ $\alpha \in \mathbb R \cup\{\infty,-\infty\}$: there is a sequence $(a_{n})_{n}\subset\mathbb R^{2}$ such that $a_{n}\to0$, and $f(a_{n})\to \alpha$.

(c) Show the partial differentiability of $f$ on $\mathbb R^{2}$.

Following ideas:

(a) Since we are in a metric space, sequential continuity implies continuity:

Using $a, b \in \mathbb R-\{0\}$, such that $x_{1}\to a$ and $x_{2}\to b$,

We find that $\lim_{(x_{1},x_{2})\to(a,b)}f(x_{1},x_{2})=\lim_{(x_{1},x_{2})\to(a,b)}\frac{x_{1}x_{2}^{2}}{\vert\vert x\vert\vert_{2}^{4}}=\lim_{(x_{1},x_{2})\to(a,b)}\frac{x_{1}x_{2}^{2}}{(|x_{1}|^{2}+|x_{2}|^{2})^{2}}$

and $\{(\cdot)^{2},|\cdot|,id\}$ are all continuous functions on $\mathbb R$, so therefore I can say:

$\lim_{(x_{1},x_{2})\to(a,b)}\frac{x_{1}x_{2}^{2}}{(|x_{1}|^{2}+|x_{2}|^{2})^{2}}=\frac{ab^{2}}{(|a|^{2}+|b|^{2})^{2}}=f(a,b)$ and therfore the function is sequentially continuous on $\mathbb R^{2}-\{0\}$, which implies continuity on the same space. Any complaints on this proof?

(b) I am a bit stumped on this one: the only nullsequence I can think of that usually helps in situations like these is: $(\frac{1}{n})_{n\in\mathbb N}$ or even $(0)_{n \in \mathbb N}$ but none of this will help in finding $\alpha$.

(c) Using the differential quotient, I want to show partial differentiability in $(0,0)$:

First in $x_{1}$ direction:

$\lim_{h\to0}\frac{f(0+he_{1})-f(0)}{h}=\lim_{h\to0}\frac{f(h,0)}{h}=\frac{\frac{h0^{2}}{(|h|^{2}+|0|^{2})^{2}}}{h}=0=\frac{d}{dx_{1}}f(0,0)$

and we find partial differential in $x_{2}$ direction in analagous fashion.

In terms of $\frac{d}{dx_{k}}f(x_{1},x_{2})$, $k \in \{1,2\}$ when $x \neq 0$, I am a bit unsure because using the differential limit gets very messy.

Any help is greatly appreciated.

Answer 1

for (b), try $x_1=1/n^2, x_2=1/n$ to get 1, and add constants to change the limit.

for (c), I think you should just derive it as usual function in $\mathbb R$, twice. once, as $x_1$ is the argument, and $x_2$ is a paramter, and vice versa. Notice that you don't get a number, but a matrix.

Answer 2

For $(b)$ transform the limit to polar coordinates

$$\lim_{(x_{1},x_{2})\to(0,0)}\frac{x_{1}x_{2}^{2}}{(x_{1}^{2}+x_{2}^{2})^{2}} = \lim_{r\to 0} \frac{r\sin\phi \cdot r^2\cos^2\phi}{r^4} = \lim_{r\to 0}\frac1r \sin\phi\cos^2\phi$$

so to get $\alpha$ you need to approach $(0,0)$ along the polar curve $$r(\phi) = \frac1\alpha \sin\phi\cos^2\phi$$

$(c)$ is correct.

For $(d)$ no need to compute the partial derivatives by definition, just use differentiation rules:

$$\frac{\partial}{\partial x_1} f(x_1, x_2) = \frac{\partial}{\partial x_1}\left(\frac{x_{1}x_{2}^{2}}{(x_{1}^{2}+x_{2}^{2})^{2}}\right) = \frac{x_2^2(x_1^2+x_2^2)^2-x_1x_2^2\cdot 2(x_1^2+x_2^2)\cdot 2x_1}{(x_1^2+x_2^2)^4} = \frac{x_2^2 (-3 x_1^2 + x_2^2)}{(x_1^2 + x_2^2)^3}$$

and similarly

$$\frac{\partial}{\partial x_2} f(x_1, x_2) = \frac{\partial}{\partial x_2}\left(\frac{x_{1}x_{2}^{2}}{(x_{1}^{2}+x_{2}^{2})^{2}}\right) = \frac{2 x_1 x_2 (x_1^2 - x_2^2)}{(x_1^2 + x_2^2)^3}$$