How could I approach $\lim_{n\to\infty}\left(\frac{1+\cos\left(\frac{1}{2^{n}}\right)}{2}\right)^n$? So, about the following limit:
$$\lim_{n\to\infty} \left( \frac{1+\cos(\frac{1}{2^{n}})}{2} \right)^n$$
I tried several things to evaluate it, namely looking at it as $\cos(\frac{1}{2^{n+1}})^{2n}$ instead or as $\exp(2n \cdot \ln({\cos(\frac{1}{2^{n+1}})})$ and then trying to show that the limit of $n\cdot\ln({\cos(\frac{1}{2^{n+1}})})$ is $0$ (for example using L'Hopital's rule), but I haven't been very successful (though it is possible I gave up too early). I'm just not sure how to approach this cosine-exponential combo. I believe the limit is $1$, so things like the root test weren't very helpful in this case, either.
I'd really appreciate a direction/hint, a full solution, or anything inbetween.
 A: \begin{align*}
\left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\leq\left(\dfrac{1+1}{2}\right)^{n}=1.
\end{align*}
Now $\cos u\geq 1-u^{2}/2$ for small $u\geq 0$, then
\begin{align*}
\left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\geq\left(1-\dfrac{1}{2^{2(n+1)}}\right)^{n}\rightarrow 1.
\end{align*}
A: Continuing where you left off with L'Hopital's:
$$L = \lim_{n \to \infty} n\ln\left(\cos\left(\frac{1}{2^{n+1}}\right)\right) = \lim_{n \to \infty} \frac{\ln(\cos(2^{-n-1}))}{\frac{1}{n}} = \lim_{n \to \infty} \frac{-\tan(2^{-n-1})\cdot2^{-n-1}\ln2}{-\frac{1}{n^2}}$$
Supposing the limits are finite, we can break up the limit over multiplication like this:
$$ \ln2 \lim_{n \to \infty} \tan(2^{-n-1}) \lim_{n \to \infty} \frac{n^2}{2^{n+1}}$$
The first limit is $0$ by the continuity of $\tan$, and the second is easily shown to be $0$ by repeatedly applying L'Hopital's.
So, our original limit is $e^{2L} = e^0 = 1.$
A: You were very close to the right track. Write
$$y=\left(\frac{1+\cos \left(\epsilon ^n\right)}{2}\right)^n\implies\log(y)=n\log\left(\frac{1+\cos \left(\epsilon ^n\right)}{2}\right)$$ Now, using the series expansion of $\cos(t)$ for small angles (or equivalents
$$\cos \left(\epsilon ^n\right)=1-\frac 12 \epsilon ^{2n}+\cdots$$
$$\frac{1+\cos \left(\epsilon ^n\right)}{2}=1-\frac 14 \epsilon ^{2n}+\cdots$$
$$\log\left(\frac{1+\cos \left(\epsilon ^n\right)}{2}\right)=-\frac 14 \epsilon ^{2n}+\cdots$$ But $\epsilon=\frac 12$; so
$$\log(y) \to 0 \implies y \to 1$$
