- #1

Hernaner28

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## Homework Statement

It says:

[tex] \displaystyle f:{{\mathbb{R}}^{2}}\to \mathbb{R}[/tex]

[tex] \displaystyle f\left( x,y \right)=\left\{ \begin{align}

& 1\text{ if 0<y<}{{\text{x}}^{2}} \\

& 0\text{ in other cases} \\

\end{align} \right.[/tex]

Show that all the directional derivatives about (0,0) exist but f is not continuous in (0,0).

## Homework Equations

Directional derivative:

[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]

## The Attempt at a Solution

I write the equation for the directionar derivative respect to a generic v=(a,b) about the origin:

[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]

That is:

[tex] \displaystyle =\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( ha,hb \right)}{h}[/tex]

I should arrive that the limit does not depend on v=(a,b). But I'm stuck here. Which is the value of [tex] \displaystyle f\left( ha,hb \right)[/tex] 1 or 0? It clearly depends on the vector v. If I get close to the origin by the y-axis then [tex] \displaystyle f\left( 0,hb \right)=0[/tex].

Thanks!

PD: I hope I'm not breaking the rules. I've received several infractions which I'm not going to discuss here. It's not my intention.

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