Expert Answer to "Evaluatinglimn→∞n22−2cos⁡(360∘n)"

Alexis Alexander

Answered question

2022-04-04

Evaluating

lim_{n \to \infty} \frac{n}{2} \sqrt{2 - 2 \cos (\frac{360^{\circ}}{n})}

Karsyn Wu

Beginner2022-04-05Added 17 answers

Notice,

lim_{n \to \infty} \frac{n}{2} \sqrt{2 - 2 \cos (\frac{2 π}{n})}

= lim_{n \to \infty} \frac{n}{2} \sqrt{2 (2 \sin^{2} (\frac{π}{n}))}

= lim_{n \to \infty} \frac{2 n}{2} \sin (\frac{π}{n})

= lim_{n \to \infty} \frac{\sin (\frac{π}{n})}{\frac{1}{n}}

= π lim_{n \to \infty} \frac{\sin (\frac{π}{n})}{(\frac{π}{n})}

Let

\frac{π}{n} = t \Rightarrow t \to 0

n \to \infty

= π lim_{t \to 0} \frac{\sin t}{t}

= π \times 1 = π

Alisha Chambers

Beginner2022-04-06Added 10 answers

Here, we will use the expansion of the cosine as

\cos x = 1 - \frac{1}{2} x^{2} + O (x^{4})

(1)
Letting

x = \frac{2 π}{n}

in (1) yields

2 - 2 \cos (\frac{2 π}{n}) = \frac{4 π^{2}}{n^{2}} + O (n^{- 4})

Finally, we have

\frac{n}{2} \sqrt{2 - 2 \cos (\frac{2 π}{n})} = \frac{n}{2} \sqrt{\frac{4 π^{2}}{π^{2}} + O (n^{- 4})}

= \frac{n}{2} \frac{2 π}{n} (1 + O (n^{- 2}))

= π + O (n^{- 3}) \to π

as expected!

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