Prove that $$f(x,y)=\begin{cases}\dfrac{x^2}{y}&\text{if $(x,y)\neq(x,0)$},\\f(x,0)=0\end{cases}$$ is derivable for all direction in $(0,0)$ but it is not differentiable at $(0,0)$.

I have 3 questions:

  1. I think that the function can be translated to $$f(x,y)=\begin{cases}\dfrac{x^2}{y}&\text{if $(x,y)\neq(x,0)$},\\\color{red}0&\color{red}{\text{if $(x,y)=(x,0)$}},\end{cases}$$ right?

To prove that has directional derivative in all direction in $(0,0)$ we need to prove that the following limit exists: $$\lim_{h\to0}\frac{f(ah,bh)-f(0,0)}{h},$$ where $\check{v}=(a,b)\in\Bbb R^2$ and $a^2+b^2=1$.

Indeed, $$\lim_{h\to0}\frac{f(ah,bh)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{(ah)^2}{bh}-0}{h}=\lim_{h\to0}\frac{a^2h^2}{bh^2}=\frac{a^2}{b}=\begin{cases}\frac{a^2}{b}&\text{if $b\neq0$},\\\color{blue}0&\color{blue}{\text{if $b=0$}},\end{cases}$$ thus the limit exists for all direction.

  1. The text in $\color{blue}{\text{blue}}$ is correct because we know that $b$ goes in $y$-direction, and $f(x,y)=0$ if $y=0$?

To prove that $f$ is not differentiable at $(0,0)$ we can study the continuity of $f$ at $(0,0)$: $f(0,0)=0$, but $$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{x^2}{y}\underbrace{=}_{(*)}\underset{y=x^2}{\lim_{x\to0}}\frac{x^2}{x^2}=1,$$ where in $(*)$ we have taken the curve of level $1$ of $f$, thus $f$ is not continuous at $(0,0)$. Hence, it is not differentiable at $(0,0)$.

  1. Can we take the curve of level $1$ just "imposing" $\frac{x^2}{y}=1$ i.e. $y=x^2$, or conversely, we need to prove that for all $(x,y)\in E^*(0,0)\cap\{(x,y)\in\Bbb R^2\mid(x,y)\neq(0,0)\}$, it is $y=x^2$?



Question 1: Yep, you can rewrite it like this. In fact, I strongly prefer this to the way it was originally written; I don't like how $f(x, y)$ appears on the right hand side of the $=$ sign. You could replace the condition $(x, y) = (x,0)$ with $y = 0$ too.

Question 2: Yes, that would be right. Again, I have some issues with how this is written out.

$$\lim_{h\to0}\frac{f(ah,bh)-f(0,0)}{h}\color{red}=\lim_{h\to0}\frac{\frac{(ah)^2}{bh}-0}{h}=\lim_{h\to0}\frac{a^2h^2}{bh^2}=\frac{a^2}{b}\color{red}=\begin{cases}\frac{a^2}{b}&\text{if $b\neq0$},\\0&\text{if $b=0$},\end{cases}$$

I don't like the $=$ signs highlighted in red. It's not true in general that $f(ah, bh) = \frac{(ah)^2}{bh}$; it makes the implicit assumption that $b \neq 0$. Similarly, the second highlighted $=$ is a second mistake, designed to correct the previous one, sneaking the $b = 0$ case back into the equation. You'll notice that the $b = 0$ case is not proven, merely asserted. No wonder you need to ask for clarification!

Question 3: You can just look at the curve $y = x^2$ (or parameterised, $(x,y) = (t, t^2)$). It is a continuous curve, passing through $(x, y) = (0, 0)$. If $f$ is continuous, then we'd expect $f(t, t^2)$ to be a continuous function of $t$, and in particular, $f(t, t^2) \to f(0, 0)$ as $t \to 0$. As this is not the case, $f$ is not continuous.

  • 1
    $\begingroup$ Wow, great analysis!! I have doubts in your answer (2). If the first $=$ in red is wrong, then my whole study is wrong because we cannot deduce the RHS. So, what would you write to prove the derivability? $\endgroup$ – manooooh May 25 at 1:18
  • 1
    $\begingroup$ @manooooh Well, it's wrong in the sense that it's only right most of the time. If $b = 0$, then $$\lim_{h \to 0} \frac{f(ah, bh) - f(0, 0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0,$$ not $\frac{a^2}{b}$ (which is undefined). The equalities bury the $b = 0$ case, and reintroduce it at the end, without actually dealing with it properly. To prove it properly, write the equalities as written up to $\frac{a^2}{b}$, but first assume $b \neq 0$. Then do the $b = 0$ case (as I did in this comment). Finally, you can put them together to get the final equality. $\endgroup$ – Theo Bendit May 25 at 8:45

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.