# existence of the directional derivative of a function

Let $$f:\mathbb R^2\rightarrow\mathbb R,(x,y)↦\begin{cases}1&(\exists z\in\mathbb R\setminus\{0\}:(x,y)=(z,z ^2)\\0 &(\textrm{else})\end{cases}$$

$f$ is obviously not differentiable in $(0,0)$.

But what about any directional derivative $v$ in $(0,0)$? So I have to consider $\lim_{h\rightarrow0}\frac{f(hv)}h$. But why does this limit always exists?

If $f(hv)=0$ for all $h$ the limit is obviously 0 and also for a finite number of $f(hv)=1$.

In the other case $f(hv)=f(h(v_1,v_2))=1$ for $hv_2=h^2v_1^2\Leftrightarrow v_2=hv_1^2$ and now I am stuck. Does this case exists? And what about the limit of $\frac{f(hv)}h$ ? Wouldn't I get "$\frac10$" ?

$\lim_{h\rightarrow 0} \frac{f(hv_1,hv_2)}{h}=0$,to prove it we show that for all $\epsilon>0$, there exists $\delta>0$ such that for all $h$ : $0<|0-h|<\delta$ implies $|\frac{f(hv_1,hv_2)}{h}-0|<\epsilon$. Setting $\delta=|\frac{v_2}{2v_1^2}|$ should make it work.

(Thanks for Sheldoor for pointing out a mistake in my previous solution)

• two questions: 1) why do you have a $h'$? 2) how do you get $\delta$? I've tried to estimate the absolute value but didn't get along. And won't we get the problem of $f(-,-)=1$ for the limit?
– user81109
Jun 28, 2013 at 23:46
• @jul8 I was applying the defintion of a limit. As for h' I will write h instead. I did not mean to write it
– Amr
Jun 28, 2013 at 23:47
• okay, I see what you mean. but how did you get $\delta$ ?
– user81109
Jun 28, 2013 at 23:56
• @jul8 Is your question why does this delta work ? or you understand that it works and you want to ask how I got it ?
– Amr
Jun 28, 2013 at 23:58
• both ;) I still have problems with the points $(x,y)=(z,z^2)$ for the limit.
– user81109
Jun 28, 2013 at 23:59