# Prove that $f(x,y)$ is derivable for all direction in $(0,0)$ but it is not differentiable at $(0,0)$

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 $$y=x^2$$?

Thanks!

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.
$$\lim_{h\to0}\frac{f(ah,bh)-f(0,0)}{h}\color{red}=\lim_{h\to0}\frac{\frac{(ah)^2}{bh}-0}{h}=\lim_{h\to0}\frac{a^2h^2}{bh^2}=\frac{a^2}{b}\color{red}=\begin{cases}\frac{a^2}{b}&\text{if b\neq0},\\0&\text{if b=0},\end{cases}$$
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.
• 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? – manooooh May 25 at 1:18
• @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. – Theo Bendit May 25 at 8:45