How do I justify the integral over a limit

Question

Define $$I_n := \int_0^{\frac{1}{2^n}}(1+\cos{x})^ndx$$

Does the limit as $n \to \infty$ of $I_n$ exists? and in what way?

Usually I justify the convergence of the integral by assuming that $lim \int_D f_n(x)dx = \int_Df(x)dx$ when the function uniformly converges in $D$, however this obviously won't work here.

How do I find the value of the integral? if it even exists at all?

Answer 1

For $0\le x\le1 $, we have $\;1-\dfrac{x^2}2\le \cos x\le 1$, so $\;\Bigl(2-\dfrac{x^2}2\Bigr)^n\le(1+\cos x)^n\le 2^n$.

On the other hand, Bernoulli's inequality says that (note $-\frac{x^2}4>-1$) $$\Bigl(2-\dfrac{x^2}2\Bigr)^{\!n}=2^n\Bigl(1-\dfrac{x^2}4\Bigr)^{\!n}\ge 2^n\Bigl(1-\dfrac{nx^2}4\Bigr),$$ so that, on the interval $\Bigl[0,\frac1{2^n}\Bigr]$, $$ 2^n\biggl(1-\dfrac{n}{4^{n+1}}\biggr)\le (1+\cos x)^n\le 2^n,$$ and finally by the mean value inequality, $$\frac{2^n}{2^n}\biggl(1-\dfrac{n}{4^{n+1}}\biggr)=1-\dfrac{n}{4^{n+1}}\le\int_0^{\frac{1}{2^n}}(1+\cos{x})^ndx\le \frac{2^n}{2^n}=1.$$ The squeezing principle then shows the limit of the integrals is equal to $1$.

Answer 2

HINT:

Note that for $x\in[0,1/2^n]$, we have

$$2-2^{-(2n+1)}\le 1+\cos(x)\le 2$$