Suppose we have i.i.d samples \(X_1, \dots, X_n\) and our goal is to estimate \(\theta\). Brushing aside technicalities, let us define a differentiable function \(\rho: \mathcal{X} \times \Theta \to \mathbb{R}\) where \(\psi(x, \theta) = \frac{\partial \rho(x, \theta)}{\partial \theta}\). Then estimator \(\hat{\theta} = \underset{\theta}{\text{argmax}} \ \sum_{i = 1}^n \rho(X_i, \theta)\) is an \(M\)-estimator. The true parameter value \(\hat{\theta}_0\) we wish to estimate can be given as \(\theta_0 = \underset{\theta}{\text{argmax}} \ \mathbb{E}[\rho(X, \theta)]\). Note that these expectations are over our samples, and we do not assume any knowledge of the distribution of \(X\). By definition of the \(M\)-estimator, we have \(\sum_{i} \psi(X_i, \hat{\theta}) = 0\) and so using a Taylor Series expansion:
\[\small 0 = \sum_{i} \psi(X_i, \hat{\theta}) \approx \sum_{i} \psi(X_i, \theta_0) + \sum_{i} \psi'(X_i, \theta_0) \implies \sqrt{n}[\hat{\theta} - \theta_0] \approx \frac{\sqrt{n} \sum_{i} \psi(X_i, \theta_0)}{ - \sum_{i} \psi'(X_i, \theta_0)} = \frac{ \sum_{i} \psi(X_i, \theta_0) / \sqrt{n}}{ - \sum_{i} \psi'(X_i, \theta_0) / n}\]This is nearly identical logic to this derivation of the Sandwich Asymptotic Variance for MLEs in model misspecification scenarios. This is indeed because the MLE is a case of the general \(M\)-estimator. Using that derivation, we arrive at:
\[\sqrt{n}[\hat{\theta} - \theta_0] \overset{d}{\to} \mathcal{N}(0, V^{-1}WV^{-1})\]where \(V = \mathbb{E}[\psi'(X, \theta_0)]\) and \(W = \mathbb{E}[\psi(X, \theta_0)^2]\).