0

I got a function:

$f(x,y)= \begin{cases}\frac{x^3-3\,x\,y^2}{x^2+y^2} &(x,y) \neq 0\\ 0&(x,y) = 0 \end{cases}$

substitution polar coordinates ($x = r\,\cos(\varphi)$, $y = r\,\sin(\varphi)$) I get:

$f(x,y) = r\,\cos(3\,\varphi)$. Supposly this has to be continuous, because:

for $r_n \xrightarrow{} 0$ the function approaches zero: $\underset{n \to \infty}{\lim} r_n \,\cos(3\varphi) = 0$.

However: If I chose $(x_n,y_n) = (\frac{1}{n},\frac{1}{n})\xrightarrow{} (0,0)$, I also get:

$\underset{n\to\infty}{\lim} f(x_n,y_n) = \underset{n\to\infty}{\lim} -\frac{1}{n} = 0 $. Does this also proof continuity at $0$?

Leon
  • 1,117

1 Answers1

1

$f(x,y)=\frac{x^3-3xy^2}{x^2+y^2}$

Use the change of variables given by $x=r\cos \theta,y=r \sin \theta$ notice that When $(x,y)\to(0,0)$ then $r\to 0$.Remplazing our change of variables and symplifying the expression you should get.

$$\lim_{r \to 0} r\left(cos^3 \theta -3\sin^2\theta \cos \theta\right)=0$$

Which is unique for different $\theta$, and then $f$ is continuous at $(0,0)$

  • Your answer ist wrong ! You wrote $\lim_{r \to 0} r^3 \cos^{3}\theta -3 \sin^2 \theta \cos \theta$, but it should read $\lim_{r \to 0} r( \cos^{3}\theta -3 \sin^2 \theta \cos \theta).$ – Fred Feb 17 '21 at 14:28
  • 1
    Thanks for the correction I made a mistake –  Feb 17 '21 at 14:55