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I am struggling to show if this does or does not exist:

$$\lim_{(x,y) \to (0,0)}\frac{x^3y^2}{x^4+y^6}$$

I usually have no problems with limits like these. This one is throwing me through a loop because the power of $x$ is larger than the power of $y$ in the numerator, but the opposite is true in the denominator. So alot of the usual tricks for doing this kind of limit aren't working.

I tried approaching along all sorts of lines of the form $y=kx^a$ and I'm always getting that the limit is $0$. However I am struggling to prove this using the limit definition or the squeeze theorem.

The triangle inequality, $2|x||y|\leq x^2 + y^2$, and other tricks aren't working for me.

Thanks for the help.

NOTE: We have not covered polar coordinates, so the solution shouldn't have to use that.

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  • $\begingroup$ See here for a general case. $\endgroup$ – Arnaud D. Dec 3 '19 at 16:19
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Try using AM-GM to show that $y^6+x^4\geq C\cdot y^2\cdot x^{\frac{8}{3}}$ by writing $x^4=\frac{1}{2}x^4+\frac{1}{2}x^4$.

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  • $\begingroup$ Great answer (1+). $\endgroup$ – Jacky Chong Oct 16 '16 at 23:17
  • $\begingroup$ That did it! Thanks. $\endgroup$ – A guy Oct 16 '16 at 23:22
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$$\lim_{(x,y) \to (0,0)}\frac{x^3y^2}{x^4+y^6}$$ Consider $y=kx^a; a,k\ne 0$: $$ \frac{k^2x^{3+2a}}{x^4+k^6x^{6a}}=k^2\frac{x^{2a-1}}{1+k^6x^{6a-4}} $$ We can check that for all: $a>\frac{2}{3},\quad \frac{2}{3}>a>\frac{1}{2},\quad \frac{1}{2}>a, \quad a=\frac{1}{2}$ and $a=\frac{2}{3}$: $$ \lim_{x\to 0}k^2\frac{x^{2a-1}}{1+k^6x^{6a-4}}=0,\quad a\in\mathbb{R}/\{0\},k\in\mathbb{R}/\{0\} $$ which implies, that for every $C^{\infty}$ function $y(x)$ initial limit exists and equal to $0$.

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