Asymptotic normality of a quadratic form 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 prove (or disprove) 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.
 A: There is some difficulty when using Delta method.
It's more convenient to derive it by hand.
By law of large number, $\hat{C}\xrightarrow{P} C$. Hence $\hat{C} +\gamma_n I\xrightarrow{P} C$. Apply Slutsky's theorem, we have
$$\sqrt{n}(\hat{C} +\gamma_n I)^{-1/2}(\bar{X}-\mu)\xrightarrow{d}\mathcal{N} (0,C^{-1}).$$
By continuous mapping theorem, we have
$$
{n}(\bar{X}-\mu)^T
(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)\xrightarrow{d}\sum_{i=1}^p \lambda_i^{-1}(C)\chi^2_1.
$$
Hence
$$
\sqrt{n}(\bar{X}-\mu)^T
(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)\xrightarrow{P}0.
$$
By Slutsky's theorem, we have
$$
\sqrt{n}\mu^T(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)\xrightarrow{d}\mathcal{N} (0,\mu^T C^{-2}\mu).
$$
Combining the above two equality yields
\begin{align}
&\sqrt{n}\big(\bar{X}^T
(\hat{C} +\gamma_n I)^{-1}\bar{X}-\mu^T
(\hat{C} +\gamma_n I)^{-1}\mu\big)
\\
=
&\sqrt{n}\Big((\bar{X}-\mu)^T
(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)-2\mu^T(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)\Big)
\\
=&-2\sqrt{n}\mu^T(\hat{C} +\gamma_n I)^{-1}(\bar{X}-\mu)+o_P(1)
\\
\xrightarrow{d}&\mathcal{N} (0,4\mu^T C^{-2}\mu).
\end{align}
The remaining task is to deal with
$$
\sqrt{n}
\Big(
\mu^T
(\hat{C} +\gamma_n I)^{-1}\mu-\mu^T
(C)^{-1}\mu
\Big).
$$
Unfortunately, this term dose NOT converges to $0$.
The behavior become complicated and depends on the third and fourth moments.
To be simple, below we assume $X_i$ are normal distributed and $\gamma_n=o(n^{-1/2})$.
It's a standard result that 
$$
\sqrt{n}(\hat{C}-C)\xrightarrow{d}C^{1/2} W C^{1/2},
$$
where $W$ is a symmetric random matrix with diagonal elements as $\mathcal{N}(0,2)$ and off diagonal elements as $\mathcal{N}(0,1)$.
Thus,
$$
\sqrt{n}(\hat{C}+\gamma_n I-C)\xrightarrow{d}C^{1/2} W C^{1/2},
$$
By matrix taylor expantion $(I+A)^{-1}\sim I-A+A^2$, we have
\begin{align}
&\sqrt{n}\Big((\hat{C} +\gamma_n I)^{-1}-
C^{-1}\Big)=
\sqrt{n}C^{-1/2}\Big(\big(C^{-1/2}(\hat{C} +\gamma_n I)C^{-1/2}\big)^{-1}-I\Big)C^{-1/2}\\
=&\sqrt{n}C^{-1}\Big(\hat{C} +\gamma_n I-C\Big)C^{-1}+O_P(n^{-1/2})
\xrightarrow{d}C^{-1/2} W C^{-1/2}.
\end{align}
Thus,
$$
\sqrt{n}
\Big(
\mu^T
(\hat{C} +\gamma_n I)^{-1}\mu-\mu^T
(C)^{-1}\mu
\Big)\xrightarrow{d}\mu^T C^{-1/2} W C^{-1/2}\mu
\sim N(0,(\mu^T C^{-1}\mu)^2).$$
Thus,
\begin{align}
\sqrt{n}\big(\bar{X}^T
(\hat{C} +\gamma_n I)^{-1}\bar{X}-\mu^T
C^{-1}\mu\big)
\xrightarrow{d}\mathcal{N} (0,4\mu^T C^{-2}\mu+(\mu^T C^{-1}\mu)^2).
\end{align}
