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Background on how I got the intuition:
I have recently learnt about two variable limits, and my professor gave us a tip that whenever we have two homogenous polynomials in the denominator and numerator, with $(0,0)$ being the only problematic point, then we can decide if the limit exists and equal to zero or DNE, based on the powers. and we got some examples:

$\lim_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^4}$, DNE because the power in the denominator is $4$ and numerator $3$.
$\lim_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^2}=0$, because the power in the numerator is bigger than the denominator.
Basically if the power of the numerator is higher than the power of the denominator, the limit exists and equal zero, else it DNE.
The Limit that confused me:

$\lim_{(x,y)\to (0,0)}\frac{x^4y^13}{x^8+y^{18}}=0$, but power in denominator is $18$ and power in numerator is $17$, so based on the tip he told us, the limit shouldn't exist.. (and I can see that $(0,0)$ is the only problematic point in the denominator).

What am I missing?

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The examples you give do not involve homogeneous polynomials. Your professor's tip does not apply.

Meanwhile $$ (x^4 \pm y^9)^2 \geq 0 $$ $$ x^8 \pm 2x^4 y^9 + y^{18} \geq 0 $$ $$ x^8 + y^{18} \geq 2 x^4 |y^9 | $$ $$ \frac{2 x^4 |y|^9}{x^8 + y^{18}} \leq 1 $$ $$ \frac{2 x^4 |y|^{13}}{x^8 + y^{18}} \leq y^4 $$

I went through Lagrange multipliers for $$ f(x,y) = \frac{x^a y^b}{x^{2c} + y^{2d}} $$ with $a,b,c,d $ positive integers.

When $ad-2cd+bc=0$ the ratio $f$ is bounded but has no limit.

When $ad-2cd+bc<0$ there is no limit and the ratio is not bounded.

When $ad-2cd+bc>0$ the limit is $0$

Your question has $a=4, b=13, c=4, d=9$ so that $ad-2cd+bc=36- 72 +52 =16 > 0 $ so the limit is zero.

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  • $\begingroup$ Thanks for the answer, so I should never judge a limit by the powers? is there any intuition you would recommend once approaching a limit to know if to try to prove or disprove? $\endgroup$ – Pwaol Apr 29 at 16:01
  • $\begingroup$ You used the word homogeneous in your question but you don't appear to know what it means. For inhomogeneous polynomials, the method of Lagrange Multipliers tells us what to expect for a limit, if not otherwise clear. $\endgroup$ – Will Jagy Apr 29 at 16:05
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    $\begingroup$ Your professors rule was for homogeneous polynomials and its just fine for those. So if each term of the polynomials have equal power it is good. Here the denominator is $x^8 + y^{18}$ and the powers of the terrms are $8$ and $18$. They are not equal so the rule does not apply. But if it were $x^{18} + y^{18}$ or $x^{8} + y^8$ it would. $\endgroup$ – fleablood Apr 29 at 16:05
  • $\begingroup$ I still can't understand, probably that I didn't get what homogeneous means, here's some more example he gave: $\frac{y^4x^3}{x^4y^2+x^2y^4+x^6+y^6}$ the limit goes to zero when $x,y$ go to zero, the powers aren't the same here, is it with the word homogeneous that I'm not understanding? I just have the basic idea of homogeneous systems of equations, and connecting it from there $\endgroup$ – Pwaol Apr 29 at 16:36
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    $\begingroup$ @Pwaol for this new example the top and bottom really are homogeneous: $4+2 = 2+4 = 6 = 6$ If the numerator is changed to $x^3 y^3$ we get another bounded fraction, which can be checked by polar coordinates readily. Then the extra power of $y$ makes the limit bounded times $y$ or zero. One of the things you should practice is using polar coordinates in these problems. $\endgroup$ – Will Jagy Apr 29 at 16:40

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