Let $k$ be a real number. Define $f\colon\Bbb{R}^2\to\Bbb{R}$ by $$f(x,y)=\begin{cases}\dfrac{x^ky}{x^2+y^2}&\text{if $(x,y)\neq(0,0)$},\\k-2&\text{if $(x,y)=(0,0)$}.\end{cases}$$
- Find the value of $k$ such that $f$ is continuous at $(0,0)$.
- For the value of $k$ found in part (1), determine whether $f$ is differentiable at $(0,0)$.
So, I have solved the part (1) and found that the $k=2$.
Then both partial derivatives $f_x(0,0)$ and $f_y(0,0)$ of the function $f(x,y)$ equal $0$.
I know that if the partial derivatives exist and continuous at $(0,0)$, then the function is differentiable at $(0,0)$.
However, I also know that we can check the differentiability using this formula:
Definition. Let $f\colon X\to\Bbb{R}$ where $X\subset\Bbb{R}^2$ is open, and let $\mathbf{a}\in X$. Suppose $f_x(\mathbf{a}),f_y(\mathbf{a})$ exist. We say that $f$ is differentiable at $\mathbf{a}$ if $$\lim_{\mathbf{x}\to\mathbf{a}}\frac{f(\mathbf{x})-[f(\mathbf{a})+f_x(\mathbf{a})(x_1-a_1)+f_y(\mathbf{a})(x_2-a_2)]}{\|\mathbf{x}-\mathbf{a}\|}=0.$$
And using, this formula I get that it is not continuous at $(0,0)$ (I have used $x=r\cos\alpha$ and $y = r\sin\alpha$ substitution to check it.)
Can you please tell me which approach is right, and is the function differentiable at $(0,0)$ if $k=2$?
Best Answer
We need to check whether
$$0 = \lim_{(a_1,a_2) \to (0,0)} \frac{f(a_1,a_2)}{\|(a_1, a_2)\|} = \lim_{(a_1,a_2) \to (0,0)} \frac{a_1^2a_2}{(a_1^2+a_2^2)^{3/2}}$$
If we approach $0$ along the line $y = x$ in the first quadrant, we get
$$\frac{f(t,t)}{\|(t,t)\|} = \frac{t^3}{(2t^2)^{3/2}} = \frac1{2^{3/2}}$$
which doesn't converge to $0$ as $t \to 0^+$ and therefore the above limit is not $0$.
We conclude that $f$ is not differentiable at $(0,0)$.