How could I approach \lim_{n \to \infty} (\frac{1+\co

Mabel Breault 2022-01-02 Answered
How could I approach
\(\displaystyle\lim_{{{n}\to\infty}}{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\)
I tried several things to evaluate it, namely looking at it as \(\displaystyle{{\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}^{{{2}{n}}}}\) instead or as \(\displaystyle{\exp{{\left({2}{n}\cdot{\ln{{\left({\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}\right)}}}\right.}}}\) and then trying to show that the limit of \(\displaystyle{n}\cdot{\ln{{\left({\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}\right)}}}\) is 0 (for example using L'Hopital's rule), but I haven't been very successful

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Expert Answer

Jim Hunt
Answered 2022-01-03 Author has 1929 answers
\(\displaystyle{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\leq{\left({\frac{{{1}+{1}}}{{{2}}}}\right)}^{{n}}={1}\)
Now \(\displaystyle{\cos{{u}}}\geq{1}-{\frac{{{u}^{{2}}}}{{{2}}}}\) for small \(\displaystyle{u}\geq{0}\), then
\(\displaystyle{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\geq{\left({1}-{\frac{{{1}}}{{{2}^{{{2}{\left({n}+{1}\right)}}}}}}\right)}^{{n}}\to{1}\)
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movingsupplyw1
Answered 2022-01-04 Author has 5411 answers
Continuing where you left off with L'Hopital's:
\(\displaystyle{L}=\lim_{{{n}\to\infty}}{n}{\ln{{\left({\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}\right)}}}=\lim_{{{n}\to\infty}}{\frac{{{\ln{{\left({\cos{{\left({2}^{{-{n}-{1}}}\right)}}}\right)}}}}}{{{\frac{{{1}}}{{{n}}}}}}}=\lim_{{{n}\to\infty}}{\frac{{-{\tan{{\left({2}^{{-{n}-{1}}}\right)}}}\cdot{2}^{{-{n}-{1}}}{\ln{{2}}}}}{{-{\frac{{{1}}}{{{n}^{{2}}}}}}}}\)
Supposing the limits are finite, we can break up the limit over multiplication like this:
\(\displaystyle{\ln{{2}}}\lim_{{{n}\to\infty}}{\tan{{\left({2}^{{-{n}-{1}}}\right)}}}\lim_{{{n}\to\infty}}{\frac{{{n}^{{2}}}}{{{2}^{{{n}+{1}}}}}}\)
The first limit is 0 by the continuity of \(\displaystyle{\tan{}}\), and the second is easily shown to be 0 by repeatedly applying L'Hopital's.
So, our original limit is \(\displaystyle{e}^{{{2}{L}}}={e}^{{0}}={1}\)
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Vasquez
Answered 2022-01-08 Author has 8850 answers

You were very close to the right track. Write
\(y=(\frac{1+\cos(\epsilon^n)}{2})^n \Rightarrow \log(y)=n \log(\frac{1+\cos(\epsilon^n)}{2})\)
Now, using the series expansion of cos(t) for small angles (or equivalents
\(\cos(\epsilon^n)=1-\frac12 \epsilon^{2n}+\ldots\)
\(\frac{1+\cos(\epsilon^n)}{2}=1-\frac14 \epsilon^{2n}+\ldots\)
\(\log(\frac{1+\cos(\epsilon^n)}{2})=-\frac14 \epsilon^{2n}+\ldots\)
But \(\epsilon=\frac12\) ; so
\(\log(y) \to 0 \Rightarrow y \to 1\)

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