Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $

Question

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.

Answer 1

Actually, that function isn't even bounded near $(0,0)$ and therefore, the limit at that point doesn't exist. You can check that$$\frac{8x^2(-x^3+x^6)^3}{x^9+(-x^3+x^6)^3}=\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}$$and that$$\lim_{x\to0}\left|\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}\right|=\infty.$$

Answer 2

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