Theorem (Central limit): Let $\x_1, \x_2, \ldots$ be iid random vectors with mean $\bmu$ and variance $\bS$. Then
\[\sqrt{n}(\bar{\x}_n - \bmu) \inD \Normal(\zero, \bS).\]Proof: Let $\varphi$ denote the characteristic function of $\x - \bmu$ (which does not require a subscript since the random vectors are iid), and let $\varphi_n$ denote the characteristic function of $\sqrt{n}(\bar{\x}-\bmu) = \tfrac{1}{\sqrt{n}}\sum_i (\x_i - \bmu)$. The numbers in parentheses refer to the numbered properties of characteristic functions.
\[\begin{alignat*}{2} \varphi_n(t) &= \varphi(t/\sqrt{n})^n &\hspace{4em}& (2), (4), \textnormal{ independence} \\ &= [\varphi(\zero) + \nabla\varphi(0) \tp \t/\sqrt{n} + \tfrac{1}{2n} \t \tp \nabla^2 \varphi(\tilde{\t}/\sqrt{n}) \t ]^n && \href{taylor-expansion.html}{\textnormal{Taylor expansion}}; (6); \href{random-vector.html}{\Ex \norm{\x}^2 < \infty} \\ &= [1 + \tfrac{1}{2n} \t \tp \nabla^2 \varphi(\tilde{\t}/\sqrt{n}) \t ]^n && (1), (5), \Ex(\x_i - \bmu) = \zero \\ &\to \exp\left\{ -\tfrac{1}{2} \t \tp \bS \t \right\} && \href{exponential-limit.html}{\textnormal{Exponential limit}}; (6) \end{alignat*}\]Therefore,
\[\sqrt{n}(\bar{\x} - \bmu) \inD \Normal(\zero, \bS)\]by the continuity theorem and (8).