Show f is not differentiable at x=0

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Question (C) {22 marks}

(c) {22 markes} Let $$ f({\bf x}) = \begin{cases} \dfrac{x_1 x_2^2}{x_1^2 + x_2^2} & : {\bf x} \ne {\bf 0} \\[1ex] 0 & : {\bf x} = {\bf 0} \end{cases} $$

i. Compute $\partial f / \partial x_1$ and $\partial f/\partial x_2$ for ${\bf x} \ne (0,0)$.

ii. Compute $\partial f / \partial x_1$ and $\partial f/\partial x_2$ for ${\bf x} = (0,0)$.

iii. Using the preceding results, show that $$\lim\limits_{t\to 0} \dfrac{\partial f}{\partial x_2}(t,t) \ne \dfrac{\partial f}{\partial x_2}(0,0)$$

iv. Show that $f$ is not differentiable at ${\bf x} = {\bf 0}$

Given: $\lim\limits_{t\to 0} \frac{\partial f}{\partial x_2}(t,t) \ne \frac{\partial f}{\partial x_2}(0,0)$ Show $f$ is not differentiable at ${\bf x}={\bf 0}$.

I have completed all of the parts to this question, and after I have obtained the result from part 3, I am wondering what answer would be sufficient for part 4, I feel that part 4 is implicit from part 3.

Answer for part 4)

Part 3 shows that the solution $\frac{\partial f}{\partial x_2}$ does not exist, therefore $f$ is not differentiable at ${\bf x}={\bf 0}$.

My answer seems incomplete, but I dont know what else to add.

Any hints or help would be greatly appreciated.

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Here is one way to show part 4):

$\displaystyle \lim_{(h,k)\to(0,0)}\frac{|f(0+h,0+k)-f(0,0)-f_{x_{1}}(0,0)h-f_{x_{2}}(0,0)k|}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to(0,0)}\frac{\left\vert\frac{hk^2}{h^2+k^2}\right\vert}{\sqrt{h^2+k^2}}=$

$\displaystyle \lim_{(h,k)\to(0,0)}\frac{|h|k^2}{(h^2+k^2)^{3/2}} \ne0\;\;$ since along $h=k$,

$\;\;\displaystyle \lim_{(h,k)\to(0,0)}\frac{|h|k^2}{(h^2+k^2)^{3/2}}=\lim_{k\to0}\frac{|k|^3}{(2k^2)^{3/2}}=\frac{1}{2^{3/2}}$.