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

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$$\lim_{x\to0}\left|\frac{8 \left(x^3-1\right)^3}{x \left(x^6-3 x^3+3\right)}\right|=\infty.$$


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

Tags:

Limits

Real Analysis

Rational Functions

Multivariable Calculus

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