Theorem (Delta method): Let $g: \real^d \to \real^k$ such that $\nabla g$ is continuous in a neighborhood of $\bmu \in \real^d$ and suppose $\sqrt{n}(\x_n - \bmu) \inD \x$. Then
\[\sqrt{n}(g(\x_n) - g(\bmu)) \inD \nabla g(\bmu) \tp \x.\]Proof: We begin by noting that if $\sqrt{n}(\x_n - \bmu) \inD \x$, we must have $\x_n \inP \bmu$; showing this is a good exercise. Now, letting $N_r(\bmu)$ denote the neighborhood of $\bmu$ where $\nabla g$ is continuous,
\[\begin{alignat*}{2} & \x_n \in N_r \implies \exists \tilde{\x}_n: g(\x_n) = g(\bmu) + \nabla g(\tilde{\x}_n) \tp (\x_n - \bmu) &\hspace{3em}& \href{taylor-expansion.html}{\textnormal{Taylor expansion}} \\ \tag*{$\tcirc{1}$} & \phantom{\x_n \in N_r \implies} \sqrt{n} \{g(\x_n) - g(\bmu)\} = \nabla g(\tilde{\x}_n) \tp [\sqrt{n}(\x_n - \bmu)] \\ & \Pr\{\x_n \in N_r\} \to 1 && \x_n \inP \bmu \\ & \tilde{\x}_n \inP \bmu && \tilde{\x}_n \in LS(\bmu, \x_n); \x_n \inP \bmu \\ & \nabla g(\tilde{\x}_n) \inP \nabla g(\bmu) && \href{continuous-mapping-theorem.html}{\text{CMT}}; \nabla g \textnormal{ is continuous} \\ & \sqrt{n}(g(\x_n) - g(\bmu)) \inD \nabla g(\bmu) \tp \x && \href{slutsky.html}{\text{Slutsky}}; \tcirc{1}; \sqrt{n}(\x_n - \bmu) \inD \x \end{alignat*}\]It is worth noting that while applying a Taylor expansion requires conditions, these conditions only need to be met with probability tending to 1 in order to establish convergence in distribution. This is a consequence of the asymptotic equivalence lemma.
Typically, the distribution involved is the normal distribution as a consequence of the central limit theorem, leading directly to this corollary.
Corollary (Delta method): Let $g: \real^d \to \real^k$ such that $\nabla g$ is continuous in a neighborhood of $\bmu \in \real^d$ and suppose $\sqrt{n}(\x_n - \bmu) \inD \Normal(\zero, \bS)$. Then
\[\sqrt{n}(g(\x_n) - g(\bmu)) \inD \Normal(\zero, \nabla g(\bmu) \tp \bS \nabla g(\bmu)).\]