lim-x-x-1-cos-pi-x-

Question Number 196534 by cortano12 last updated on 27/Aug/23

\lim_{x \to \infty} x \sqrt{1 - \cos (\frac{π}{x})} = ?

Answered by Mathspace last updated on 27/Aug/23

1 - c o s u \sim \frac{u^{2}}{2} (u \to o) \Rightarrow

1 - c o s (\frac{π}{x}) \sim \frac{π^{2}}{2 x^{2}} (x \to \infty) a n d

\sqrt{1 - c o s (\frac{π}{x})} \sim \frac{π}{2 ∣ x ∣} \Rightarrow

x \sqrt{1 - c o s (\frac{π}{x})} \sim \frac{π}{2} s (x)

s (x) = 1 i f x > 0 a n d s (x) = - 1

i f x < 0 s o

{l i m}_{x \to \infty} x \sqrt{1 - c o s (\frac{π}{x})} = \bar{+} \frac{π}{2}

Answered by Frix last updated on 27/Aug/23

0 < t = \sqrt{1 - \cos \frac{π}{x}} \Leftrightarrow x = \frac{π}{\cos^{- 1} (1 - t^{2})}

x \to \infty \Rightarrow t \to 0

\lim_{x \to \infty} x \sqrt{1 - \cos \frac{π}{x}} =

= \lim_{t \to 0^{+}} \frac{π t}{\cos^{- 1} (1 - t^{2})} = [l^{'} H \hat{o p i t a l}]

= \lim_{t \to 0^{+}} \frac{π}{\frac{2}{\sqrt{2 - t^{2}}}} = \frac{π}{\sqrt{2}}

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