2
$\begingroup$

Let $k$ be a real number. Define $f\colon\Bbb{R}^2\to\Bbb{R}$ by $$f(x,y)=\begin{cases}\dfrac{x^ky}{x^2+y^2}&\text{if $(x,y)\neq(0,0)$},\\k-2&\text{if $(x,y)=(0,0)$}.\end{cases}$$

  1. Find the value of $k$ such that $f$ is continuous at $(0,0)$.
  2. For the value of $k$ found in part (1), determine whether $f$ is differentiable at $(0,0)$.

(Image that replaced text.)

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 $$\lim_{\mathbf{x}\to\mathbf{a}}\frac{f(\mathbf{x})-[f(\mathbf{a})+f_x(\mathbf{a})(x_1-a_1)+f_y(\mathbf{a})(x_2-a_2)]}{\|\mathbf{x}-\mathbf{a}\|}=0.$$

(Image that replaced text.)

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 $k=2$?

$\endgroup$
  • $\begingroup$ Welcome to math.SE!! $\endgroup$ – manooooh Dec 26 '19 at 21:18
  • 1
    $\begingroup$ Yes, $k=2$ is correct. You mean to say that $f$ is not differentiable at $(0,0)$? Please show the details of how you showed that limit is not $0$. $\endgroup$ – Ted Shifrin Dec 26 '19 at 22:20
1
$\begingroup$

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 $(0,0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.