Question Number 196534 by cortano12 last updated on 27/Aug/23
$$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\sqrt{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{x}}\right)}\:=? \\ $$
Answered by Mathspace last updated on 27/Aug/23
$$\mathrm{1}−{cosu}\sim\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\left({u}\rightarrow{o}\right)\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\sim\frac{\pi^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }\:\left({x}\rightarrow\infty\right)\:{and} \\ $$$$\sqrt{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)}\sim\frac{\pi}{\mathrm{2}\mid{x}\mid}\:\Rightarrow \\ $$$${x}\sqrt{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)}\sim\frac{\pi}{\mathrm{2}}{s}\left({x}\right) \\ $$$${s}\left({x}\right)=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:{and}\:{s}\left({x}\right)=−\mathrm{1} \\ $$$${if}\:{x}<\mathrm{0}\:{so} \\ $$$${lim}_{{x}\rightarrow\infty} {x}\sqrt{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)}=\overset{−} {+}\frac{\pi}{\mathrm{2}} \\ $$
Answered by Frix last updated on 27/Aug/23
![0<t=(√(1−cos (π/x))) ⇔ x=(π/(cos^(−1) (1−t^2 ))) x→∞ ⇒ t→0 lim_(x→∞) x(√(1−cos (π/x))) = =lim_(t→0^+ ) ((πt)/(cos^(−1) (1−t^2 ))) = [l′Ho^ pital] =lim_(t→0^+ ) (π/(2/( (√(2−t^2 ))))) =(π/( (√2)))](https://www.tinkutara.com/question/Q196566.png)
$$\mathrm{0}<{t}=\sqrt{\mathrm{1}−\mathrm{cos}\:\frac{\pi}{{x}}}\:\Leftrightarrow\:{x}=\frac{\pi}{\mathrm{cos}^{−\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$${x}\rightarrow\infty\:\Rightarrow\:{t}\rightarrow\mathrm{0} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\sqrt{\mathrm{1}−\mathrm{cos}\:\frac{\pi}{{x}}}\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\pi{t}}{\mathrm{cos}^{−\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\pi}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$