Question Number 175147 by infinityaction last updated on 20/Aug/22
![if x is a real number in [0,1] then the value of lim_(m→∞ ) lim_(n→∞) [1+cos^(2m) (n!πx)]](https://www.tinkutara.com/question/Q175147.png)
$$\:\:\:\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\:\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\underset{{m}\rightarrow\infty\:} {\mathrm{lim}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2}{m}} \left({n}!\pi{x}\right)\right]\: \\ $$
Answered by floor(10²Eta[1]) last updated on 21/Aug/22
![since cos(x) is limited then −1≤lim_(n→∞) cos(n!πx)≤1 ⇒0≤lim_(n→∞) cos^(2m) (n!πx)≤1 lim_(x→∞) a^x =0 where 0≤a<1 ⇒lim_(m→∞) [lim_(n→∞) cos^(2m) (n!πx)]=0 when cos(n!πx)≠1 cos(n!πx)=1⇒n!πx=0⇒x=0 if x≠0⇒lim_(m→∞) lim_(n→∞) [1+cos^(2m) (n!πx)]=1 if x=0⇒lim_(m→∞) lim_(n→∞) [1+cos^(2m) (n!πx)]=2](https://www.tinkutara.com/question/Q175151.png)
$$\mathrm{since}\:\mathrm{cos}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{limited}\:\mathrm{then} \\ $$$$−\mathrm{1}\leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}cos}\left(\mathrm{n}!\pi\mathrm{x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}cos}^{\mathrm{2m}} \left(\mathrm{n}!\pi\mathrm{x}\right)\leqslant\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}a}^{\mathrm{x}} =\mathrm{0}\:\mathrm{where}\:\mathrm{0}\leqslant\mathrm{a}<\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\left[\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}cos}^{\mathrm{2m}} \left(\mathrm{n}!\pi\mathrm{x}\right)\right]=\mathrm{0}\:\mathrm{when}\:\mathrm{cos}\left(\mathrm{n}!\pi\mathrm{x}\right)\neq\mathrm{1} \\ $$$$\mathrm{cos}\left(\mathrm{n}!\pi\mathrm{x}\right)=\mathrm{1}\Rightarrow\mathrm{n}!\pi\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{x}\neq\mathrm{0}\Rightarrow\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2m}} \left(\mathrm{n}!\pi\mathrm{x}\right)\right]=\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{x}=\mathrm{0}\Rightarrow\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2m}} \left(\mathrm{n}!\pi\mathrm{x}\right)\right]=\mathrm{2} \\ $$