The limit of the perimeter of the regular n-g | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) regular $n$-gon inscribed in a circle of radius $R$ has a perimeter given by $P_n = 2nR \sin(\frac{\pi}{n})$.As $n 	o \infty$,we use the limit $\

Vedclass · Accepted Answer

$2\pi R$

Vedclass · Answer

(A) regular $n$-gon inscribed in a circle of radius $R$ has a perimeter given by $P_n = 2nR \sin(\frac{\pi}{n})$.As $n 	o \infty$,we use the limit $\lim_{x 	o 0} \frac{\sin x}{x} = 1$.Substituting $x = \frac{\pi}{n}$,as $n 	o \infty$,$x 	o 0$.Thus,$\lim_{n 	o \infty} P_n = \lim_{n 	o \infty} 2nR \sin(\frac{\pi}{n}) = \lim_{n 	o \infty} 2\pi R \frac{\sin(\pi/n)}{\pi/n} = 2\pi R(1) = 2\pi R$.Therefore,the perimeter approaches the circumference of the circle,which is $2\pi R$.

The limit of the perimeter of the regular $n$-gons inscribed in a circle of radius $R$ as $n \to \infty$ is

Similar Questions

If $\operatorname{Lim}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}=p$,then $96 \log _e p$ is equal to . . . . . .

$\lim _{x \rightarrow 0} \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}} = $

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} = $

$\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$ is equal to

$\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module