Question about proof of a characteristic function

Question

Let $X_k = 1$ with probability $0.5$ and $X_k = -1$ with probability 0.5, and let $X_k$ be independent random variables $k = (1,2,...,n)$.

I was able to prove that the characteristic function $\Omega_X (\omega)$ of the random variable $\sum_{i=1}^n \frac{X_i}{\sqrt{n}}$ equals $\left(\cos{\frac{\omega}{\sqrt{n}}}\right)^n$ without too much trouble.

However I'm struggling to prove the next part of the question, which asks me to prove that the limit of that characteristic function approaches $e^{-\frac{\omega^2}{2}}$as n approaches infinity.

There is a hint provided which says that I should take the logarithm of the characteristic function and then apply L'Hopital's rule, but I'm struggling at the LHOP part. I'm sure it's a weakness in my calc skills, but I seem to be stuck unfortunately. Any help would be much appreciated!

Problem is from Schaum's Outline of Probability and Statistics, 4th edition. P 3.74.

Answer 1

$$\lim_{t \to 0} \frac {\ln (\cos (\omega \sqrt t)} t$$ $$=\lim_{t \to 0} \frac {-\omega \sin (\omega (\sqrt t)\frac 1 2 t^{-1/2}} {{\cos (\omega (\sqrt t)}}$$ $$=-\frac {\omega^{2}} 2$$ using the fact that $\frac {\sin x} x \to 1$ as $ x \to 0$. Put $t=\frac 1n$ to see that the required limit is $e^{-\frac {\omega^{2}} 2}$.

Answer 2

Recall that $$\cos z=\sum^\infty_{k=0}\frac{(-1)^kz^{2k}}{(2k)!}=1-\frac{z^2}{2}(1+g(z))$$ where $\lim_{z\rightarrow0}g(z)=0$.

Also, recall that for any sequence of complex numbers $c_n\xrightarrow{n\rightarrow\infty}c$, $$\lim_{n\rightarrow\infty}\Big(1+\tfrac{c_n}{n}\Big)^n=e^c$$

Then $$\begin{align} \Big(\cos\frac{\omega}{\sqrt{n}}\Big)^n&=\left(1-\frac{\omega^2}{2n}\Big(1+g\big(\frac{\omega^2}{n}\big)\Big)\right)^n\\ &=\left(1+\frac{-\tfrac{\omega^2}{2}(1+g\big(\tfrac{\omega^2}{n}\big))}{n}\right)^n\xrightarrow{n\rightarrow\infty}e^{-\tfrac{\omega^2}{2}} \end{align}$$