Check if limit exists and its value

Question

How to check if this limit exists: $$\lim_{(x,y)\rightarrow (0,0)}{ \frac{x^4y^4}{(x^2 + y^4)^3}}$$

Can I convert it to polar form ? $$\frac{r^8 (\cos^4 (\theta)\sin^4(\theta)}{r^6 (\cos^6 (\theta) + r^6 \sin^{12} (\theta))} = \frac{r^2}{\cos^6(\theta) + 0} = 0$$

I mistrust the denominator on the leftmost part of the equation. $(a+b)^3 \ne a^3 + b^3.$ — David K, Jul 03 '14 at 14:32

score 2 · Accepted Answer · answered Jul 03 '14 at 14:29

2

The limit does not exist. If we appoach along either $x=0$ or $y=0$ we get a limit of zero, if we approach along $x=t^2$, $y=t$ we get a limit of $\frac{1}{8}$.

answered Jul 03 '14 at 14:29

Rene Schipperus

39,526

How did you knew what values for x and y choose? ($t$,$t^2$) – Giannis Foulidis Jul 03 '14 at 23:13
You try to make the top and bottom homogenous of the same degree, plus someone else mentioned it in their comments... – Rene Schipperus Jul 03 '14 at 23:18

score 1 · Answer 2 · answered Jul 03 '14 at 14:20

1

Consider $x(t)=t^2$, $y(t)=t$, so that the fraction behaves, as $t \to 0$, like a nonzero constant.

answered Jul 03 '14 at 14:20

Siminore

35,136

Check if limit exists and its value

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