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

$\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2} = $

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module