The limit of the interior angle of a regular | vedclass

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

Question

(A) The interior angle of a regular polygon with $n$ sides is given by the formula: $	heta_n = \frac{(n-2) 	imes 180^{\circ}}{n}$.We want to find th

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) The interior angle of a regular polygon with $n$ sides is given by the formula: $	heta_n = \frac{(n-2) 	imes 180^{\circ}}{n}$.We want to find the limit as $n ightarrow \infty$:$\lim_{n ightarrow \infty} 	heta_n = \lim_{n ightarrow \infty} \frac{(n-2) 	imes 180^{\circ}}{n} = \lim_{n ightarrow \infty} (1 - \frac{2}{n}) 	imes 180^{\circ}$.As $n ightarrow \infty$, the term $\frac{2}{n} ightarrow 0$.Therefore, the limit is $(1 - 0) 	imes 180^{\circ} = 180^{\circ} = \pi 	ext{ radians}$.

The limit of the interior angle of a regular polygon of $n$ sides as $n \rightarrow \infty$ is

Similar Questions

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

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}} \right)$ equals

If $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$,then $m(m-1)=$

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

$\mathop {Lim}\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$ when $n$ is an integer:

Vedclass Test Series

Exam Paper Generator

Online Exam Module