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$ 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.

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  • $\begingroup$ L'hôpital's doesn't work for multiple variables (at least we were taught it doesn't). Should I derive for x or y? It doesn't make sense $\endgroup$ – SlowerPhoton Oct 4 '17 at 7:08
  • $\begingroup$ this trick may help $\endgroup$ – mattapow Oct 4 '17 at 7:11
  • $\begingroup$ Yes, it did. Included this approach in answers. $\endgroup$ – SlowerPhoton Oct 4 '17 at 7:24
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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 $(0,0)$.

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  • $\begingroup$ Remark: This is the usual trick if you got functions of this form :) $\endgroup$ – Yaddle Oct 4 '17 at 7:25
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    $\begingroup$ Small side note: The case $x = y = 0$ is irrelevant to the limit. Limits by definition ignore whatever happens on the specific point they approach. Of course, you have to check that the limit as $(x, y)\to (0,0)$ agrees with the function value at $(0,0)$ for continuity. $\endgroup$ – Arthur Oct 4 '17 at 7:25
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$x = r\cos \theta$, $y = r\sin \theta$

instead of $(x,y) \rightarrow (0,0)$ I can now use $r\rightarrow0$

$$\begin{align} \lim_{r\to0} \frac{r^3\cos^3\theta + r^3\sin^3\theta}{r^2\cos^2\theta + r^2\sin^2\theta} &=\, \lim_{r\to0} \frac{r (\cos^3\theta + \sin^3\theta)}{\cos^2\theta + \sin^2\theta} \\ &=\,0 \end{align}$$

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    $\begingroup$ This is not effective. You have to show the convergence is independent of your choice of $\theta$. $\endgroup$ – Vim Oct 4 '17 at 7:26
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    $\begingroup$ @Vim It is: Include an additional step wherte you note that the final denominator is equal to $1$, along with $-1\leq \sin^3\theta, \cos^3\theta\leq 1$, and this becomes quite clear. $\endgroup$ – Arthur Oct 4 '17 at 7:28
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    $\begingroup$ @Arthur yeah. Adding this will make this answer more clear. $\endgroup$ – Vim Oct 4 '17 at 7:30
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    $\begingroup$ And in fact to prove the 2D limit, the limit needs to have the same value not just for any fixed $\theta$, but even if $\theta$ varies as a function of $r$ (which is still true here). $\endgroup$ – aschepler Oct 4 '17 at 9:50
  • $\begingroup$ As ashepler said, this answer is not correct as stated. The limit must be independent of the path taken to the limit point; for example, the result must also be the same for the path $y=x^2$ and $x\rightarrow0$. It is, as the other answer shows, but that doesn't follow from this argument. $\endgroup$ – tomsmeding Oct 4 '17 at 10:51

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