$ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3+y^3}{x^2+y^2} $ seems pretty hard to analyse.
2 Answers
$\begingroup$For $x,y \in \mathbb R \setminus \{0\}$ we have
$$\vert f(x,y)\vert = \left\vert\frac{x^3 + y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2+ y^2} \right\vert + \left\vert \frac{y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2} \right\vert + \left\vert \frac{y^3}{y^2}\right\vert = \vert x \vert + \vert y \vert \to 0$$ for $(x, y) \to (0, 0)$. If $x = y = 0$ we have $f(x,y) = 0$. Thus it follows that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$. Therefore we can deduce that $f$ is continuous at $(0,0)$.
$\endgroup$ 2 $\begingroup$$x = r\cos \theta$, $y = r\sin \theta$
instead of $(x,y) \rightarrow (0,0)$ I can now use $r\rightarrow0$
$$\begin{align} \lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2\cos^2\theta + r^2\sin^2\theta} &=\, \lim_{r\to0} \frac{r (\cos^3\theta + \sin^3\theta)}{\cos^2\theta + \sin^2\theta} \\ &=\,0 \end{align}$$
$\endgroup$ 6
