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Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.

Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 \,.$$

Approaching this limit from $y=0$, $x=0$, $y=x$, $y=x^2$ etcetera all yields 0 as value, so my proposal is that this limit is indeed 0.

I have been able to solve most similar limits so far by finding some convergent upper bound for the absolute limit, but with this one the difference between the numerator and the denominator is so small I can't find anything to fit inbetween. For example, $(x,y)\to(0,0)$, $$ \left| \frac{x^2 y}{x^2 + y^2} \right| \le \left| \frac{(x^2 + y^2)y}{x^2 + y^2} \right| \to 0 \,. $$

Also, Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? and Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. contain some helpful hints.

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PHPirate
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  • @Αδριανός Thanks for the different suggestion! It seems to work fine, at least it's a systematic way, which is great! – PHPirate Jun 21 '16 at 14:34
  • @Αδριανός I think you accidentally left out the factor $r$ in the polar formulas. – StackTD Jun 21 '16 at 14:34
  • @StackTD You're right, I even read over that haha. – PHPirate Jun 21 '16 at 14:34
  • Hard to let $r \to 0$ when there is no $r$ ;o). – StackTD Jun 21 '16 at 14:35
  • Ah right sorry about that; that would certainly make a difference! – KR136 Jun 21 '16 at 14:36
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    I believe switching to polar coordinates also works and is quite effective. Use the normal substitutions $x=r\cos(\theta)$ and $y=r\sin(\theta)$ then change the limit to the single variable one as $r$ goes to $0$. Note that it doesn't matter how you approach the origin, so the theta values don't matter to a certain extent under the limit. – KR136 Jun 21 '16 at 14:37
  • @StackTD That is quite important to the limit ;) Here is the amended comment. – KR136 Jun 21 '16 at 14:38
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    Through this index I found another question about the same limit. However, the other question specifically asks for $\varepsilon$-$\delta$ proof. – Martin Sleziak Jun 21 '16 at 18:43
  • These two questions are very closely related, too: http://math.stackexchange.com/questions/1822811/continuity-of-fracx3y2x4y4-at-0-0 and http://math.stackexchange.com/questions/856098/proving-fracx3y2x4y4-is-continuous – Martin Sleziak Jun 21 '16 at 18:47
  • @MartinSleziak Oh, that's a useful index! The other question about the same limit is indeed not really the same, but the hints given would have made me able to solve this one, the same with the other two questions mentioned. I am pretty curious why those other questions didn't came up in the first two pages on google... A google search just brings me back to my own question. For that reason I'll not delete the question but edit it a little, I hope that's the right thing to do. – PHPirate Jun 23 '16 at 09:45
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    See also the general case at https://math.stackexchange.com/questions/66226/multivariable-limit-proof-lim-x-y-rightarrow-0-0-frac-leftx-righta – Arnaud D. Dec 03 '19 at 16:28

3 Answers3

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You have: $$(x^2-y^2)^2 \ge 0 \Rightarrow x^4+y^4 \ge 2x^2y^2$$ So also: $$x^2y^2 \le x^4+y^4$$ Which means: $$\left| \frac{x^2y^3}{x^4+y^4} \right| \le \left| \frac{x^2y^3}{x^2y^2} \right| = |y|$$

StackTD
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    Thanks, this is what I was searching for! Always seems to simple in hindsight... @GFauxPas I googled it, it seems to be the same written in a little different form. – PHPirate Jun 21 '16 at 14:20
  • You're welcome; the inequality $x^4+y^4 \ge 2x^2y^2$ often comes in handy with these kind of problems. – StackTD Jun 21 '16 at 14:21
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In polar coordinates the expression is

$$r\frac{\cos^2t\sin^3t}{\cos^4 t + \sin^4 t}.$$

The denominator in this fraction has a positive minimum; thus the fraction is a bounded function of $t,$ and the $r$ in front guarantees a limit of $0.$

zhw.
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Hint 1: $$\frac{x^2 y^3}{x^4+y^4}=\frac{y}{(x/y)^2+(y/x)^2}. $$

Hint 2: $f(u)=u^2+u^{-2}$ has an absolute minimum.

Vincenzo Oliva
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  • I agree that it works, but I like StackTD's answer better because I don't know how I could have thought of this... – PHPirate Jun 21 '16 at 14:24
  • @PHPirate: I just divided top and bottom by $x^2y^2$. Hadn't seen the other answer, in fact this is another way to get the same inequality. – Vincenzo Oliva Jun 21 '16 at 14:24
  • I saw that, I was talking about thinking of the absolute minimum of $u^2 + u^{-2}$ ;) – PHPirate Jun 21 '16 at 14:32
  • @PHPirate: Oh. Well, the function is continuous on its domain, positive and goes to $\infty$ as $x\to0,\pm\infty$, hence it can be expected. :) – Vincenzo Oliva Jun 21 '16 at 14:40