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I'm trying to show that a multivariable function is continuous at the point $(0,0)$. To do so, I'd like to show the following limit:

$\lim_{(x_1, x_2) \to (0,0)} \frac{(x_1)^3(x_2) - (x_1)(x_2)^3}{(x_1)^2 +(x_2)^2}=0$

Unfortunately, I don't have much experience with limits of multiple variables. Can anyone help?

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Notice that, for $x_1^2+x_2^2 >0$ $$ \left| \frac{(x_1)^3(x_2) - (x_1)(x_2)^3}{(x_1)^2 +(x_2)^2} \right| = \left| x_1 x_2 \frac{ x_1^2 - x_2^2}{x_1^2+x_2^2} \right| \leqslant \left| x_1 x_2 \right| $$ meaning that $$ -x_1 x_2 \leqslant \frac{(x_1)^3(x_2) - (x_1)(x_2)^3}{(x_1)^2 +(x_2)^2} \leqslant x_1 x_2 $$ The limit of both bounding functions is zero, hence the limit in question is zero as well.

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Why is it okay to just take the absolute value like that? – Cal C. Apr 18 '12 at 16:53
@CalC. I have expanded the post a little. Hope this helps. – Sasha Apr 18 '12 at 16:58
Very much so. Thanks! – Cal C. Apr 18 '12 at 17:00

Changing to polar coordinates works nicely here: $$\eqalign{ \lim_{(x_1,y_1)\rightarrow(0,0)}\ {x_1^3x_2-x_1x_2^3\over x_1^2+x_2^2} &=\lim_{r\rightarrow 0^+}\ {r^4\cos^3\theta\sin\theta -r^4\cos\theta\sin^3\theta\over r^2}\, \cr &=\lim_{r\rightarrow 0^+}\ \bigl[\,r^2(\cos^3\theta\sin\theta - \cos\theta\sin^3\theta)\,\bigr] \cr &=0. } $$

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