Finding $\lim_{n\to\infty}\int_0^{\frac{1}{2^n}}(1+\cos(x))^n\,dx$ During our studies on Riemann Integrals we were given the following question in my Calculus II course:
$$ \forall n\in\mathbb{N}: \text{let } f_n(x)=(1+\cos(x))^n, f_n:[0,1]\to \mathbb{R}$$
$$\text{ Find}: \lim_{n\to\infty}\int_0^{\frac{1}{2^n}}f_n(x)\, dx $$
I have tried finding the limit using the squeeze theorem. I noticed that the function is defined on $[0,1]$, where
$\cos(x)$ is positive and decreasing. I used the following boundaries:
$$ 1\leq f_n\leq 2^n \implies \frac{1}{2^n} = \int_0^{\frac{1}{2^n}}1 \leq \int_0^{\frac{1}{2^n}}f_n(x) \leq \int_0^{\frac{1}{2^n}}2^n = 1  $$
but $ \displaystyle{\lim_{n\to\infty}}\frac{1}{2^n} = 0 $ which means I'm unable to apply the squeeze theroem.
I've tried using tighter lower boundaries such as $1+\cos(1)$ But it didn't seem to work.
I did some trial and error by calculaing the integrals manually for different values of $n$ and it does seem like $\displaystyle{\lim_{n\to\infty}}\int_0^{\frac{1}{2^n}}f_n(x) dx = 1 $
This means that somehow I should find a lower bundary which tends to 1 when $n\to\infty$.
Any ideas on how to go forward from here are welcome.
 A: You have $$(1+\cos x)^n\geq \left(2-\frac{x^2}{2}\right)^n\geq\left(2-2^{-2n-1}\right)^n$$
on $x\in[0,2^{-n}]$
and therefore:
$$\int_0^{2^{-n}}f_n(x)dx\geq\left(1-\dfrac{1}{2^{2n+2}}\right)^n\to 1.$$
A: Here is yet another slightly different answer:
Since $x\mapsto \cos(s)$ monotone decreasing in $[0,\pi/2]$ and $2^{-n}\leq \tfrac2n$ for $n>2$,
$$2^{-n}\Big(1+\cos\big(\tfrac{2}{n}\big)\Big)^n\leq 2^{-n}(1+\cos(2^{-n}))^n\leq I_n:=\int^{2^{-n}}_0(1+\cos(x))^n\,dx\leq 1$$
A simple algebraic manipulation gives
$$
a_n:=2^{-n}\left(2-\Big(1-\cos\big(\tfrac2n\big)\Big)\right)^n=\left(1-\frac{n\Big(1-\cos\big(\tfrac2n\big)\Big)}{2n}\right)^n
$$
Notice that $\frac{1-\cos\big(\tfrac2n\big)}{\tfrac2n}\xrightarrow{n\rightarrow\infty}0$. Hence $a_n\xrightarrow{n\rightarrow\infty}e^0=1$
Putting things together, we gave that $\lim_nI_n=1$.

Here we are using a well known result that states that if $c_n\xrightarrow{n\rightarrow\infty}c$, then
$$\Big(1+\frac{c_n}{n}\Big)^n\xrightarrow{n\rightarrow\infty}e^c.$$
A: Another approach:
By change of variable, the integral becomes
\begin{align*}
\int_{0}^{1}\left(\cos\left(\dfrac{y}{2^{n-1}}\right)\right)^{n}dy.
\end{align*}
It is not hard to compute that
\begin{align*}
\varphi_{n}(y)=\left(\cos\left(\dfrac{y}{2^{n-1}}\right)\right)^{n}\rightarrow 1
\end{align*}
pointwise $y\in[0,1]$.
Now we look at
\begin{align*}
F(z)=\left(\cos\left(\dfrac{y}{2^{z-1}}\right)\right)^{z}
\end{align*}
for large $z>0$. It is not hard to check that $F'(z)>0$ for large $z$, this implies that $(\varphi_{n})_{n}$ is an increasing sequence.
By Dini Theorem, $\varphi_{n}$ converges uniformly to $1$, now you can swipe the limit with the integral.
EDIT:
A useful limit here:
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\log(\cos x)}{-x^{2}}=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos x}\cdot-\sin x}{-2x}=\dfrac{1}{2},
\end{align*}
so
\begin{align*}
n\log\left(\cos\left(\dfrac{y}{2^{n-1}}\right)\right)=\dfrac{\log\left(\cos\left(\dfrac{y}{2^{n-1}}\right)\right)}{-\left(\dfrac{y}{2^{n-1}}\right)^{2}}\cdot-\left(\dfrac{y}{2^{n-1}}\right)^{2}\cdot n\rightarrow 0.
\end{align*}
A: Obviously the behaviour close to $x=0$ is relevant, so we need to check the contributions of higher order:
$$\int_0^{1/2^n} (1+\cos(x))^n \, {\rm d}x = \int_0^{1/2^n} (1+1+{\cal O}(x^2))^n \, {\rm d}x = 2^n \int_0^{1/2^n} (1+{\cal O}(x^2))^n \, {\rm d}x \\
= 2^n \left( 2^{-n} + \sum_{k=1}^n \binom{n}{k} \int_0^{1/2^n} {\cal O}(x^{2k}) \, {\rm d}x \right) = 1 + \sum_{k=1}^n \binom{n}{k} \frac{{\cal O}(4^{-nk)}}{2k+1} \, .$$
Now you can estimate the sum $$\left|\sum_{k=1}^n \binom{n}{k} \frac{{\cal O}(4^{-nk)}}{2k+1}\right| \leq C\sum_{k=1}^n \binom{n}{k} 4^{-nk} = C \left\{ (1+4^{-n})^n - 1 \right\} \rightarrow 0$$
as $n\rightarrow \infty$ for some constant $C>0$.
