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?
Asked
Active
Viewed 134 times
0
-
When $\sin(\theta )=0$, then obviously $\frac{r\cos(\theta )\sin^2(\theta )}{r^2\cos^4(\theta )+\sin^2(\theta )}=0$. – Surb Oct 31 '20 at 15:47
-
Maybe this can help. – user Oct 31 '20 at 15:55
1 Answers
0
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 $$|F(r,\theta)|\le r|\cos(\theta)|\le r$$
hamam_Abdallah
- 63,719