Let $\mathbf{x}$ be a random vector drawn from $P$. Consider a sample $\{ \mathbf{x}_i \}_{i=1}^n \stackrel{i.i.d.}{\sim} P$. Define $\bar{\mathbf{x}}_n := \frac{1}{n} \sum_{i=1}^n \mathbf{x}_i$, and $\hat{C} := \frac{1}{n} \sum_{i=1}^n (\mathbf{x}_i - \bar{\mathbf{x}}_n) (\mathbf{x}_i - \bar{\mathbf{x}}_n)^\top$. Let $\boldsymbol{\mu} := \mathbb{E}_{\mathbf{x}\sim P}[\mathbf{x}]$ and $C:=\mathrm{cov}_{\mathbf{x} \sim P}[\mathbf{x}, \mathbf{x}]$.

By the central limit theorem, assume that

$$ \sqrt{n} \big( \bar{\mathbf{x}}_n - \boldsymbol{\mu} \big) \stackrel{d}{\to} \mathcal{N}(\boldsymbol{0}, C), $$

where $C$ is a full rank covariance matrix.

Question: How do I show that

$$\sqrt{n} \big( \bar{\mathbf{x}}_n^\top (\hat{C} + \gamma_n I)^{-1} \bar{\mathbf{x}}_n - \boldsymbol{\mu}^\top C^{-1} \boldsymbol{\mu} \big) \stackrel{d}{\to} \mathcal{N}(0, v^2),$$

for some $v>0$, and for some $\gamma_n \ge 0$ such that $\lim_{n\to \infty} \gamma_n =0$? This looks simple. But I could not figure it out exactly how to show this. This is not a homework question.

My understanding is that the delta method would allow us to easily conclude

$$\sqrt{n} \big( \bar{\mathbf{x}}_n^\top C^{-1} \bar{\mathbf{x}}_n - \boldsymbol{\mu}^\top C^{-1} \boldsymbol{\mu} \big) \stackrel{d}{\to} \mathcal{N}(0, v^2),$$

or

$$\sqrt{n} \big( \bar{\mathbf{x}}_n^\top (\hat{C} + \gamma_n I)^{-1} \bar{\mathbf{x}}_n - \boldsymbol{\mu}^\top (\hat{C} + \gamma_n I)^{-1} \boldsymbol{\mu} \big) \stackrel{d}{\to} \mathcal{N}(0, v^2).$$

These are a bit different from what I want. Notice the covariance matrices in the two terms. I feel that I miss something very trivial here. Alternatively, if it makes things simpler, we can also ignore $\gamma_n$ i.e., set $\gamma_n =0$ and assume that $\hat{C}$ is invertible. Thanks.