Do nice properties of MLE still hold in Classical Linear Regression Model? For simplicity, let's assume that we have the true DGP
$$y_i= \beta_0 + \beta_1x_i + \epsilon_i$$ where $\epsilon_i \sim N(0, \sigma^2)$ $(i = 1,2,...,n)$
Assume that these following usual assumptions meet (for the classical linear regression model)

*

*$y_i= \beta_0 + \beta_1x_i + \epsilon_i$ (linearity)

*$x_i$ is deterministic

*$\epsilon_i \sim N(0, \sigma^2)$ and $Cov[\epsilon_i, \epsilon_j] = 0$ for $i \neq j$ (normality, homoskedasticity and uncorrelatedness)

If we use Maximum Likelihood Estimation Method (MLE) to estimate $\beta_0$ and $\beta_1$, is it correct to state that the nice asymptotic properties as n approaches infinity (asymptotic efficiency, asymptotic consistency, ...) still hold for the MLE estimators of $\beta_0$ and $\beta_1$?
I believe that the answer is no because of the second assumption and, for example, $MSE[\hat{\beta}_{1MLE/OLS}] = \mathbb{V}[\hat{\beta}_{1MLE/OLS}] = \frac{\sigma^2}{Dev(x)}$ which implies there is no guarantee that $MSE[\hat{\beta}_{1MLE/OLS}]$ will approach 0 as n approaches infinity (sufficient but not necessary condition for convergence in probability), but I want to know whether or not it is correct.

Requested in comments
With simple linear regression, the MLE estimators the same as those from minimising the sum of squares of the residuals.
It depends on what additional $x_i$ you include as $n$ increases.  The variance of $\hat \beta_1$ is $\frac{\sigma^2}{\sum(x_i-\bar x)^2}$ which tends towards $0$ so long as the denominator increases without bound and you do not concentrate the additional $x_i$ around the existing $\bar x$.
So for example you are correct to worry if $x_1=1$ and $x_2=-1$ and all the other $x_i=0$: you will never mitigate the impact of $\epsilon_1$ and $\epsilon_2$ on $\hat \beta_1$.
But this is not realistic: the point of expanding the sample for OLS is to see the impact of different values of $x_i$ on $y_i$.
