use polar coordinates to find the limit. hint: let (x = rcos(\theta)) and (y = rsin(\theta)), and note that…

use polar coordinates to find the limit. hint: let (x = rcos(\theta)) and (y = rsin(\theta)), and note that ((x,y)\to(0,0)) implies (r\to0). (lim_{(x,y)\to(0,0)}\frac{x^{8}+y^{8}}{x^{2}+y^{2}})

use polar coordinates to find the limit. hint: let (x = rcos(\theta)) and (y = rsin(\theta)), and note that ((x,y)\to(0,0)) implies (r\to0). (lim_{(x,y)\to(0,0)}\frac{x^{8}+y^{8}}{x^{2}+y^{2}})

Answer

Explanation:

Step1: Substitute polar coordinates

Let (x = r\cos\theta) and (y = r\sin\theta). Then (x^{2}+y^{2}=r^{2}), (x^{8}+y^{8}=r^{8}\cos^{8}\theta + r^{8}\sin^{8}\theta). The limit (\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{2}}{x^{8}+y^{8}}) becomes (\lim_{r\to0}\frac{r^{2}}{r^{8}(\cos^{8}\theta+\sin^{8}\theta)}).

Step2: Simplify the expression

(\lim_{r\to0}\frac{r^{2}}{r^{8}(\cos^{8}\theta+\sin^{8}\theta)}=\lim_{r\to0}\frac{1}{r^{6}(\cos^{8}\theta+\sin^{8}\theta)}). Since (0\leqslant\cos^{8}\theta+\sin^{8}\theta\leqslant2) (because (0\leqslant\cos^{2}\theta,\sin^{2}\theta\leqslant 1)), and as (r\to0), (\frac{1}{r^{6}}\to+\infty).

Answer:

The limit does not exist.