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How to show if the limit $$ \lim_{(x,y) \to (0,0)} \frac{x^3}{x^2 + y^2}$$ exists? I suspect that there is, as I can't find any path that would show that limit doesn't exist, and WolframAlpha also suggests that the limit is (0,0).

In general, can you recommend any tips how to learn to approach similar limit problems (fractions and polynomials like this)? edit That is, excluding the polar coordinate conversion method?

kekkonen
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    $x^2 + y^2$ in the denominator of a limit question is almost always an alarm bell which screams "switch to polar!" – Sp3000 Sep 13 '13 at 04:43
  • One of the reasons for the "switch to polar" suggestions is that you have two variables simultaneously closing in on a potential singularity; further, it is not entirely clearly defined (in most cases) what it means to have a two-variable limit. So "switching to polar" is a suggestion to create a good definition of this type of limit by approaching the limit vector in a circular fashion. – abiessu Sep 13 '13 at 04:55
  • A related technique. – Mhenni Benghorbal Sep 13 '13 at 05:39

4 Answers4

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$$ \left|\frac{x^3}{x^2+y^2}\right|\le|x| $$

rubik
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Pocho la pantera
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We have to show that $\lim_{(x,y)\to (0,0)}f(x,y)=0$, so we have to prove the following claim:

$$\forall \epsilon>0,~~\exists\delta>0,~\forall(x,y)\left(0<||(x,y)-(0,0)||<\delta\Longrightarrow\left|\frac{x^3}{x^2+y^2}-0\right|<\epsilon\right)$$ We have $$||(x,y)-(0,0)||<\delta\Longrightarrow\sqrt{x^2+y^2}<\delta\longrightarrow|x|<\delta,~|y|<\delta$$ So if we set $z=\text{max}(|x|,|y|)$, then $z<\delta$ and from this we get: $$\left|\frac{x^3}{x^2+y^2}-0\right|=\frac{|x|^3}{|x|^2+|y|^2}<\frac{z^3}{z^2}=z<\delta$$ Therefore, it is sufficient that $\delta=\epsilon$.

Mikasa
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Use polar coordinates to write the limit as

$$ \lim_{r \to 0} r \cos^3\theta $$

zodiac
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0

HINT

We can use polar coordinates or also

$$\left|\frac{x^3}{x^2+y^2}\right|\le \frac{|x|^3+|x|y^2}{x^2+y^2}$$

user
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