Studio funzione in 2 variabili

[Henry!]

Ciao a tutti!

Volevo proporvi il seguente esercizio:

Data la funzione $ f(x,y)={ ( xysen(1/(xy)); xy!=0),( 0;xy=0 ):} $

Se ne determinino i punti di continuità, derivabilità, differenziabilità.

Chiaro che lo studio va fatto in $(0,0)$

Ora, per la continuità pensavo che, qualunque sia l'argomento, $ |sen(x)|<=|x| $, da cui discende in modo immediato la continuità.

Per la derivabilità: $ { ( phi(x) = f(x,0) = 0 ),( psi(y)=f(0,y)=0 ):} $

$ phi'(0) = psi'(0)= \partial_(x)f(0,0) = \partial_(y)f(0,0)$

Da cui discende la derivabilità di f.

Per la differenziabilità, essendo $f(0,0)=0$, ottengo:

$ lim_((x,y)rarr(0,0)) (xysen(1/(xy)))/sqrt(x^2+y^2) $

che per il discorso del seno di prima diventa: $|(xysen(1/(xy)))/sqrt(x^2+y^2)|<=|(xy)/sqrt(x^2+y^2)|$

ed in coordinate polari: $|rhosenthetacostheta|<=rhorarr0$ ovviamente per $rhorarr0$

Il che implica la differenziabilità.

Volevo il vostro parere, perchè mi sembra troppo facile così

Nota: Come si mettono gli spazi bianchi in ASCII?

[Henry!]

Niente?

[javicemarpe]

The continuity doesn't follow from $|\sin(x)|\leq |x|$ but from $|x\sin(1/x)|\leq |x|$.

On the other hand, if you want to study the differentiability of the function $f$ on $(0,0)$ you have to find one (the only one) linear application $L$ such that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L[(x,y)-(0,0)]}{\|(x,y)\|}=0.$$

I'll show you the step-by-step way to do it:

This $L$ is completely determined by the partial derivatives $\partial_x f(0,0)$ and $\partial_y f(0,0)$ in the following way: $L(x,y)=\partial_xf(0,0)x+\partial_yf(0,0)y$. So the only thing you have to do is to compute these derivatives.

Now, $$\partial_x f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0,$$
and the same happens with $\partial_yf(0,0)$. Then, the only candidate to be the differential of $f$ at $(0,0)$ is the null function $L(x,y)=0$.

Let's check it. First, let us observe that the candidate to be the limit is $0$ because, when $x=0$ or $y=0$, the function is identically $0$ and so is the limit of the definition of differentiability. So, we have to check if it is true that

$$0=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{\|(x,y)\|}=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$
even when we approach the point $(0,0)$ with points $(x,y)$ such that $xy\ne 0$.

But, if $xy\ne 0$,
$$\left|\frac{f(x,y)}{\|(x,y)\|}\right|=\left|\frac{xy\sin\left(\frac{1}{xy}\right)}{\|(x,y)\|}\right|\leq \frac{|xy|}{\|(x,y)\|}\leq \frac{|x|\|(x,y)\|}{\|(x,y)\|}=|x|,$$ which goes to $0$ when $(x,y)\to 0$. When $xy=0$, the function is identically $0$, so for these points is also clear that the limit is $0$. Then, the function is differentiable at $(0,0)$ and its differential is $L(x,y)=0$.

This is the step-by-step way to do it. If you check your final argument, you'll realise that what you have is that
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\left|\frac{|\rho^2\cos\theta\sin\theta|}{\rho}\right|\leq \rho\to0.$$

Probably you wanted to say that thing. Anyways, you have to be careful with this kind of inequalities.

[Henry!]

Wow, amazing answer! Thanks a lot and yes, those inequalities are so tricky!