Does $\frac{\overline X - \mu}{S/\sqrt n}$ converge to $N(0,1)$? Suppose that $X_1,...,X_n,...$ are i.i.d. with $EX_i = \mu < \infty$ and $Var X_i = \sigma^2 < \infty$. Now, I know that by the Central Limit Theorem that
$$
\frac{\overline X - \mu}{\sigma/\sqrt n} \rightharpoonup N(0,1).
$$
Now, suppose that instead of the actual $\sigma^2$, I use the sample standard deviation $S^2 = \sum^n_{i=1}\frac{(X_i-\overline X)^2}{n-1}$. Thus, is it true that
$$
\frac{\overline X - \mu}{S/\sqrt n} \rightharpoonup N(0,1).
$$
My intuition is that this is true, but surprisingly I was not able to find easily anywhere. Sorry if this is a silly question. [
 A: The sample variance $S^2$ is consistent  for $\sigma^2$.
By the Mann-Wald (continuous mapping) theorem, since the function $g: \mathbb{R}_{\geq 0} \to \mathbb{R}$ defined by $g(x) = \sqrt{x}$ is continuous, it follows that $g(S^2) = S$ is consistent for $g(\sigma^2) = \sigma$.
Write
$$\dfrac{\overline{X} - \mu}{S/\sqrt{n}} = \dfrac{\sigma}{S} \cdot \dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}}\text{.}$$
By the CLT, $\dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}} \overset{d}{\to} \mathcal{N}(0, 1)$.
By Mann-Wald again, we have that $h: \mathbb{R}_{> 0} \to \mathbb{R}$ given by $h(x) = \dfrac{\sigma}{x}$ is continuous, hence since $S$ is consistent for $\sigma$, $h(S) = \dfrac{\sigma}{S} $ is consistent for $h(\sigma) = \dfrac{\sigma}{\sigma} = 1$.
By Slutsky's Theorem, $$\dfrac{\overline{X} - \mu}{S/\sqrt{n}} = \dfrac{\sigma}{S} \cdot \dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}} \overset{d}{\to} 1 \cdot \mathcal{N}(0, 1) = \mathcal{N}(0, 1)\text{.}$$
A: Yes, this is the Slutsky's theorem https://en.wikipedia.org/wiki/Slutsky%27s_theorem.
$$\frac{\bar{X}-\mu}{S/\sqrt{n}}=\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}}\frac{\sigma}{S}\rightarrow\mathcal{N}(0,1)\quad\text{(in distribution)}$$
because $\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\rightarrow\mathcal{N}(0,1)$ in distribution and $\frac{\sigma}{S}\rightarrow 1$ in probability. This happens with every consistent estimator.
