Application of Central Limit Theorem - Uniform Distribution The following question I found on an old exam:
Given $n$ iid random variables $X_k$, $1 \leq k \leq n$, with uniform distribution on $[-1,1]$, it is easy to compute the characteristic function of one them, namely
$$ \phi_{X_k}(t) = \sin (t).$$
By the properties of the characteristic function and by independence, we also have,
$$\phi_{\bar{X_n}}(t) = [\sin(\frac{t}{\sqrt{n}})]^n.$$
Then finally one is asked, to show that
$$ \lim_{n \rightarrow \infty}\left[\frac{\sin(\frac{t}{\sqrt{n}})}{\frac{t}{\sqrt{n}}}\right]^n \rightarrow \exp(-\frac{3t^2}{2}),$$
by using the Central Limit theorem. We are also given $\mathbb{E}(X_k^2) = \frac{1}{3}$ and the hint that if $Y \sim \mathcal{N}(0, \sigma^2)$, then $\phi_Y(t) = \exp(-\frac{(\sigma t)^2}{2}).$
This is where I've been struggling so far. Any input would be greatly appreciated :) Thanks!
 A: You have made an mistake. The CF of $X_k$ is wrong. The CF is:
\begin{align}
\phi_{X_k}(t) &= \int^1_{-1}e^{itx}\frac{1}{2}dx = \frac{1}{2} \int^1_{-1} \cos(tx) + i\sin(tx)dx \\
&= \int^1_{0}\cos(tx)dx = \frac{1}{t}\sin(tx)\big|^1_0 = \frac{\sin(t)}{t} 
\end{align}
You may know that $E[X_k]=0$. It is also given that $E[X_k^2]=\frac{1}{3}$ so $\text{Var}(X_k)=\frac{1}{3}$. The CLT tells us the following:
\begin{align}
U_n=\frac{\sum_{k=1}^nX_k}{\sqrt[]{n}} \stackrel{D}{\to} \mathcal{N}(0,1/3) 
\end{align}
Now you can check that:
\begin{align}
\phi_{U_n}(t) =\phi_{X_1}(t/\sqrt[]{n})^n = \left(\frac{\sin\left(\frac{t}{\sqrt[]{n}}\right)}{\frac{t}{\sqrt[]{n}}}\right)^n
\end{align}
And the result follows directly from the continuity property of the characteristic function.
\begin{align}
\lim_{n\to\infty} \left(\frac{\sin\left(\frac{t}{\sqrt[]{n}}\right)}{\frac{t}{\sqrt[]{n}}}\right)^n = \exp\left(-\frac{t^2}{6}\right)
\end{align}
You may see that the $3$ in the numerator must be in the denominator. So the result in question contains a typo I guess. 
