<s>[INST] Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
   \lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= r^2 \sin^2(t)$, but these attempts were unsatisfactory.
I also have reviewed the similar questions and their answers but there are difference between those functions and mine one, I think the principal difference is that the powers of the denominators are odd.
 [/INST] As suggested in the comments, let consider the following path "near" the problematic points $y=-x^3$ for which denominator vanishes:

*

*$x=t$

*$y=-t^3+t^5$
then we have
$$\frac{8 x^2 y^3 }{x^9+y^3}
=\dfrac{8 t^2 (-t^3+t^5)^3 }{t^9+(-t^3+t^5)^3}
=\frac{8 t^2(-t^{9}+3t^{11}-3t^{13}+t^{15})}{t^9+(-t^{9}+3t^{11}-3t^{13}+t^{15})}=\\
=8\frac{-t^{11}+3t^{13}-3t^{15}+t^{17}}{3t^{11}-3t^{13}+t^{15}}=8\frac{-1+3t^2-3t^{4}+t^{6}}{3-3t^{2}+t^{4}}$$
For the general strategy, see also the related:

*

*Find $\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{x+y}$
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