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

Question

I have tried to write the limit using polar coordination. but I remain with a $cos(\theta)$ in the denominator. thanks for the help

Here is a similar question, just without the factor $2$ in the denominator: How to obtain $\lim_{(x,y) \to (0,0)} (x^2·y^3)/(x^4+y^6)$. — Martin Sleziak, Jul 12 '20 at 08:53

score 3 · Answer 1 · answered Jul 11 '20 at 16:19

3

Hint

Try going along the paths $(x(t), y(t)) = (t, t)$ and $(x(t), y(t)) = (t^3, t^2)$.

answered Jul 11 '20 at 16:19

Aryaman Maithani

score 3 · Accepted Answer · answered Jul 11 '20 at 16:21

3

Along the path $y=0$

$$\lim_{(x,0)\to(0,0)} \frac{0}{x^4+0} = 0$$

but along the path $ y=x^{\frac{2}{3}}$

$$\lim_{(x,x^{\frac{2}{3}})\to(0,0)} \frac{x^4}{3x^4} = \frac{1}{3}$$

thus the limit does not exist.

answered Jul 11 '20 at 16:21

Ninad Munshi

2 Answers2