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)…

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.\nlim_(x,y)→(0,0) (x^8 + y^8)/(x^2 + y^2)

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.\nlim_(x,y)→(0,0) (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+\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\neq0)) and the numerator is a non - zero constant (1).

Answer:

(\infty)