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I need to prove that $f$ continuous at $(x, y)=(0,0)$ using a $\epsilon$-$\delta$ proof $$ f(x, y) = \begin{cases} \frac{x^3}{{x^2 + y^2}},&(x,y)\neq (0,0) \\ 0,&(x,y) = (0,0) \end{cases} $$

I'm not sure how to manipulate the function to get $\delta$

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up vote 2 down vote accepted

Hint: Use that $\displaystyle\frac{x^2}{x^2+y^2}\le 1$. Then start the proof like

Assume we are given an $\varepsilon>0$. Let $\delta:=$ ...

If the $(x,y)$ point is closer to $(0,0)$ than $\delta$, then, in particular, $|x|<\delta$, so ...

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the numerator is $x^3$ – PooperScooper Feb 21 '13 at 1:20
Exactly. That is, $x\cdot x^2$. – Berci Feb 21 '13 at 1:21
oh ok ok my bad thanks! – PooperScooper Feb 21 '13 at 1:24
$\delta:=\varepsilon$ – Berci Feb 21 '13 at 1:35

You can use polar coordinates to see the problem more clearly: $$f(r,\theta)=\frac{r^3 \cos^3(\theta)}{r^2} = r \cos^3(\theta)$$ Thus, for any $\epsilon>0$, choose $\delta=\epsilon$. If $r<\delta$: $$|f(r,\theta)-0| < |r \cos^3(\theta)-0|<|r|<\delta=\epsilon$$

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