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.