Linear combination of normal distribution with Slutsky's theorem Suppose
$$\sqrt{n} (\hat{\beta} - \beta) \overset{d}{\rightarrow} N(0, \sigma^2)$$
Then I know that for some constant $\alpha$ that a linear combination of normal is normal:
$$\sqrt{n} (\alpha\hat{\beta} - \alpha\beta) \overset{d}{\rightarrow} N(0, \sigma^2\alpha^2).$$
Now suppose that $\hat{\alpha} \rightarrow \alpha$, my question is, does
$$\sqrt{n} (\hat{\alpha}\hat{\beta} - \alpha\beta) \overset{d}{\rightarrow} N(0, \sigma^2\alpha^2)$$
by Slutsky's theorem?

Your conjecture is not true.  As a counterexample, let $\alpha = 1$, $\hat{\alpha} = 1 + n^{-1/2}$ (non-random), then $\hat{\alpha} \to \alpha$, but
\begin{align}
\sqrt{n}(\hat{\alpha}\hat{\beta} - \alpha\beta) = \sqrt{n}(\hat{\beta} - \beta) + \hat{\beta}, 
\end{align}
which, by Slutsky's theorem (note that $\sqrt{n}(\hat{\beta} - \beta) \to_d N(0, \sigma^2)$ implies that $\hat{\beta} \to_P \beta$), converges to $N(\beta, \sigma^2)$ in distribution, which is not the same as $N(0, \sigma^2)$ when $\beta \neq 0$.
