Asymptotic covariance matrix of $\bar{\pmb x}$ In a text I'm reading it says that we define
$$
\begin{align}
\bar{\pmb x}= \begin{bmatrix}\bar x_1 \\ \bar x_2 \end{bmatrix}
\end{align}
$$
And then immediately says the asymptotic covariance matrix for $\bar{\pmb x}$ is
$$
\begin{align}
\pmb V = N E\left[(\bar{\pmb x} - \pmb \mu)(\bar{\pmb x} - \pmb \mu)' \right]
\end{align}
$$
But I'm having a hard time seeing that. I'm comfortable with the covariance matrix for $\pmb x$ being $E[(\pmb x - \pmb \mu)(\pmb x - \pmb \mu)']$ but for some reason I'm having a hard time seeing how $N$ comes into the picture even though I know it gets there via the CLT.
I know that the CLT says $\bar x \sim N(\mu, \frac{\sigma^2}{N})$ but still having a hard time getting from there to the definition of the asymptotic covariance matrix.
Distributional Assumptions
$x_i$ is distributed multi-variate normal, independently, and i.i.d.

Solution
It makes sense now, the variance isn't for $\pmb x$ but for $\pmb{\bar x}$. Therefore we have:
$\bar x \sim N(\mu, \sigma^2_{\bar x})$ which when we demean and multiply by $\sqrt{N}$ we have $\sqrt{N}(\bar x-\mu) \sim N(0, N\sigma^2_{\bar x})$.

Asymptotically, $\bar{x} - \mu$ goes to zero.  So the asymptotical covariance of $\bar{x}$ refers to that of $\sqrt{n}(\bar{x} - \mu)$, which is $V$, when we assume all the regularity conditions of CLT are met.
By the way, we do not really need the distributional assumption on $x$.
Edit:  (Further details as requested by OP)
$V = Var(\sqrt{n}(\bar{x} - \mu)) = n Var(\bar{x}) = n E (\bar{x}-\mu)^2$ which is what you have in your textbook.
In the event that $x_t$ are i.i.d, then $n Var(\bar{x}) = Var(x)$.  Maybe this is what confused you?  The textbook is written down without assuming i.i.d.
