Question

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}})

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.