Question

use polar coordinates to find the limit. hint: let x = r cos(θ) and y = r sin(θ), and note that (x,y)→(0,0) implies r→0.
lim_(x,y)→(0,0) (x^8 + y^8)/(x^2 + y^2)

Explanation:

Step1: Substitute polar - coordinates

Let \(x = r\cos\theta\) and \(y = r\sin\theta\). Then \(x^{2}+y^{2}=r^{2}\), \(x^{8}+y^{8}=r^{8}(\cos^{8}\theta+\sin^{8}\theta)\). The given 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 \(0\leqslant\cos^{8}\theta+\sin^{8}\theta\leqslant1\) (because \(0\leqslant\cos^{2}\theta,\sin^{2}\theta\leqslant1\)), and as \(r\to0\), the denominator \(r^{6}(\cos^{8}\theta+\sin^{8}\theta)\to0\) (when \(r
eq0\)) and the numerator is a non - zero constant \(1\).

Answer:

\(\infty\)