 Mabel Breault

2022-01-02

How could I approach
$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{1+\mathrm{cos}\left(\frac{1}{{2}^{n}}\right)}{2}\right)}^{n}$
I tried several things to evaluate it, namely looking at it as ${\mathrm{cos}\left(\frac{1}{{2}^{n+1}}\right)}^{2n}$ instead or as $\mathrm{exp}\left(2n\cdot \mathrm{ln}\left(\mathrm{cos}\left(\frac{1}{{2}^{n+1}}\right)\right)$ and then trying to show that the limit of $n\cdot \mathrm{ln}\left(\mathrm{cos}\left(\frac{1}{{2}^{n+1}}\right)\right)$ is 0 (for example using LHopitals rule), but I havent Jim Hunt

${\left(\frac{1+\mathrm{cos}\left(\frac{1}{{2}^{n}}\right)}{2}\right)}^{n}\le {\left(\frac{1+1}{2}\right)}^{n}=1$
Now $\mathrm{cos}u\ge 1-\frac{{u}^{2}}{2}$ for small $u\ge 0$, then
${\left(\frac{1+\mathrm{cos}\left(\frac{1}{{2}^{n}}\right)}{2}\right)}^{n}\ge {\left(1-\frac{1}{{2}^{2\left(n+1\right)}}\right)}^{n}\to 1$ movingsupplyw1

Continuing where you left off with LHopitals:
$L=\underset{n\to \mathrm{\infty }}{lim}n\mathrm{ln}\left(\mathrm{cos}\left(\frac{1}{{2}^{n+1}}\right)\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(\mathrm{cos}\left({2}^{-n-1}\right)\right)}{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{-\mathrm{tan}\left({2}^{-n-1}\right)\cdot {2}^{-n-1}\mathrm{ln}2}{-\frac{1}{{n}^{2}}}$
Supposing the limits are finite, we can break up the limit over multiplication like this:
$\mathrm{ln}2\underset{n\to \mathrm{\infty }}{lim}\mathrm{tan}\left({2}^{-n-1}\right)\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{2}}{{2}^{n+1}}$
The first limit is 0 by the continuity of $\mathrm{tan}$, and the second is easily shown to be 0 by repeatedly applying LHopitals.
So, our original limit is ${e}^{2L}={e}^{0}=1$ Vasquez

You were very close to the right track. Write
$y=\left(\frac{1+\mathrm{cos}\left({ϵ}^{n}\right)}{2}{\right)}^{n}⇒\mathrm{log}\left(y\right)=n\mathrm{log}\left(\frac{1+\mathrm{cos}\left({ϵ}^{n}\right)}{2}\right)$
Now, using the series expansion of cos(t) for small angles (or equivalents
$\mathrm{cos}\left({ϵ}^{n}\right)=1-\frac{1}{2}{ϵ}^{2n}+\dots$
$\frac{1+\mathrm{cos}\left({ϵ}^{n}\right)}{2}=1-\frac{1}{4}{ϵ}^{2n}+\dots$
$\mathrm{log}\left(\frac{1+\mathrm{cos}\left({ϵ}^{n}\right)}{2}\right)=-\frac{1}{4}{ϵ}^{2n}+\dots$
But $ϵ=\frac{1}{2}$ ; so
$\mathrm{log}\left(y\right)\to 0⇒y\to 1$

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