How to derive the asymptotic distribution of t-statistic?

Question

Let ${X_n}$ be an IID sample such that ${X_i} \sim N(\mu,\sigma^2)$. When both $\mu$ and $\sigma$ are unknown, we construct $t(\hat{\mu},s)=\dfrac{\sqrt{n}(\hat{\mu}-\mu)}{s}$, where $s$ is the sample standard deviation.

The statistic $t(\hat{\mu},s)$ follows t-distribution exactly. I wish to know the asymptotic distribution of $t(\hat{\mu},s)$.

I understand that by Continuous Mapping theorem under certain conditions, if $\hat{\mu} \xrightarrow{d} X \sim N(\mu,\sigma^2/n)$ then $h(\hat{\mu}) \xrightarrow{d} h(X)$. However, in case of $t(\hat{\mu},s)$, it is a function of two random variables. So how to derive its asymptotic distribution?

I also checked Slutsky's theorem. That also requires that at least one of $\hat{\mu}$ and $s$ should converge to a constant. Now by CLT, both converges in distribution to a random variable with normal distribution. However, by combining LLN we may say that $s \xrightarrow{p} \sigma$. Would it be the right way to go?

EDIT: As pointed out in the second comment this is not a duplicate of this. Seemingly $s$ converged in probability to a constant but to a random variable in distribution. This was the main doubt that whether it was okay to ignore the convergence in distribution.

Answer 1

It is not correct that $\widehat\mu$ and $s$ fail to converge in distribution to constants, nor that CLT implies that.

Rather, what CLT says is:

\begin{align} & \frac{\widehat\mu - \mu}{\sigma/\sqrt n} \overset d \longrightarrow \operatorname N(0,1) \\[8pt] \text{and } & \sqrt{n-1} \left. \left(\dfrac{s^2}{\sigma^2}-1\right) \right/\!\!\sqrt 2 \overset d \longrightarrow \operatorname N(0,1). \end{align}

It can be shown to follow that $\widehat\mu$ and $s$ converge in distribution to $\mu$ and $\sigma$ respectively.

And convergence in distribution to a constant entails convergence in probability to that constant.