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I have tried approaching the limit from different functions like $x = 0$, $y = 0$, $y = x$, $y = x^2$, etc. But they all go to $0$, so my guess would be that the limit goes to $0$, but how do I show that the limit is definitely $0$?

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

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    $\begingroup$ How do you know $\left|\frac{x^3 - y^3}{x^2 + y^2}\right|\leqslant \frac{|x|^3}{x^2 + y^2}+ \frac{|y|^3}{x^2 + y^2}$? Is there a name for this/is there a proof of it I can read online? $\endgroup$ – mr eyeglasses May 17 '14 at 22:40
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    $\begingroup$ The triangle inequality says that $|x^3 - y^3|\leqslant|x^3|+|y^3|=|x|^3+|y|^3$. $\endgroup$ – Did May 17 '14 at 22:42
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    $\begingroup$ Can you please explain why $ \frac{|x|^3}{x^2 + y^2}+ \frac{|y|^3}{x^2 + y^2}\leqslant|x|+|y|$. $\endgroup$ – Sandeep Silwal May 20 '14 at 14:52
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    $\begingroup$ @SandeepSilwal Because $\frac{|x|^3}{x^2 + y^2}\leqslant\frac{|x|^3}{x^2}=|x|$ for instance. $\endgroup$ – Did May 20 '14 at 17:17
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There is the ever-popular method of converting the expression into polar form, which will point out quickly if the effort to find the limit suffers from "directionality":

$$ \ \frac{x^3 \ - \ y^3}{x^2 \ + \ y^2} \ \ \rightarrow \ \ \frac{r^3 \ (\cos^3 \theta \ - \ \sin^3 \theta)}{r^2} \ = \ r \ (\cos^3 \theta \ - \ \sin^3 \theta) \ . $$

This will approach zero from any direction $ \ \theta \ $ as $ \ r \ \rightarrow \ 0 \ $ .

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  • $\begingroup$ Like I say, it's "ever-popular" (which is not the same as saying that it works for everything)... $\endgroup$ – colormegone May 17 '14 at 22:46

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