If $\widehat{\sigma}^2_n\xrightarrow{P}\sigma^2$and $\sqrt{n}\left(\widehat{\mu}_n - \mu\right)/\sigma\xrightarrow{d}N(0, 1)$, then show that $\sqrt{n}\left(\widehat{\mu}_n - \mu\right)/\sqrt{\widehat{\sigma}^2_n}\xrightarrow{d}N(0, 1)$.
Is this question as simple as stating that as n tends to infinity, the sample variance also tends to the true variance and therefore the distribution must also tend to a normal one? If this is not the case, why is that so?
\begin{align} \frac{ \sqrt{n}\left(\widehat{\mu}_n - \mu\right)}{\sqrt{\widehat{\sigma}^2_n}} \end{align} The numerator converges in distribution to $\operatorname N(0,\sigma^2).$ The denominator converges in probability to $\sigma\ne0.$
So Slutksy's theorem does it.