0
$\begingroup$

The limit I need to calculate is $\lim_{(x,y)\rightarrow (0,0)}\frac{xy^{2}}{x^{4}+y^{2}}$. Using polar coordinates I get: $lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^{2}(\theta)}{r^{2}\cos^{4}(\theta)+\sin^{2}(\theta)}$. Now if $\sin(\theta)\neq 0$ then the limit is $0$. How do I handle the case where $\theta=0$ or $\theta = \pi$? And is there a better way to approach this limit?

$\endgroup$
  • $\begingroup$ When $\sin(\theta )=0$, then obviously $\frac{r\cos(\theta )\sin^2(\theta )}{r^2\cos^4(\theta )+\sin^2(\theta )}=0$. $\endgroup$ – Surb Oct 31 '20 at 15:47
  • $\begingroup$ Maybe this can help. $\endgroup$ – user Oct 31 '20 at 15:55
0
$\begingroup$

For any $ x\ne 0$ and $ y\ne 0$, we have

$$x^4+y^2>y^2>0$$

$$\frac{1}{x^4+y^2}<\frac{1}{y^2}$$

$$|\frac{xy^2}{x^4+y^2}|<|x|$$

Thus, the limit is zero.

By polar coordinates, you will find that $$|F(r,\theta)|\le r|\cos(\theta)|\le r$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.