Why does $\sqrt{n} (\bar X - \mu)/S$ have approximately a $t$-distribution? Suppose that the random variables $X_1,\ldots, X_n$ are independent and identically distributed with mean $\mu$ and variance $\sigma^2$. Let $\bar X = (X_1 + \cdots + X_n)/n$ be the sample mean and $S = ((X_1 - \bar X)^2 + \cdots + (X_n - \bar X)^2)/(n-1)$ be the sample variance.
If the random variables $X_i$ are normally distributed, then $\sqrt{n}(\bar X - \mu)/\sigma$ is normally distributed and $\sqrt{n}(\bar X - \mu)/S$ has a $t$-distribution. Moreover, even if the random variables $X_i$ are not normally distributed, then $\sqrt{n}(\bar X - \mu)/\sigma$ is approximately normally distributed, according to the central limit theorem.
Question: In this case (where the random variables $X_i$ are not normally distributed), does the central limit theorem also imply that $\sqrt{n}(\bar X - \mu)/S$ has approximately a $t$-distribution? 

This question arises in the context of constructing confidence intervals for the mean of a random variable, in the case where the variance is unknown.
The following passage appears in Ross's book Introduction to Probability and Statistics for Engineers and Scientists:

Our derivations of the $100(1-\alpha)$ percent confidence intervals
  for the population mean $\mu$ have assumed that the population
  distribution is normal. However, even when this is not the case, if
  the sample size is reasonably large then the intervals obtained will
  still be approximate $100(1-\alpha)$ percent confidence intervals for
  $\mu$. This is true because, by the central limit theorem,
  $\sqrt{n}(\bar X - \mu)/\sigma$ will have approximately a normal
  distribution, and $\sqrt{n}(\bar X - \mu)/S$ will have approximately a
  $t$-distribution.

I'm trying to understand the final statement about $\sqrt{n}(\bar X - \mu)/S$ having approximately a $t$-distribution.
 A: According to the central limit theorem $\sqrt{n}(\bar X - \mu)/\sigma$ tends to standard normal distribution as $n$ tends to infinity. But by the same token, since $s$ tends to $\sigma$ a.s., the variable $\sqrt{n}(\bar X - \mu)/s$ tends to standard normal distribution as well. Since the $t$-distribution is a sequence of distributions and not the single distribution it's hard to define what we mean by "tends" to. If we mean it in the sense of the Kolmogorov-Smirnov distance or something along these lines, we then indeed can say that it approaches the $t$-distribution. But it's only the consequence of the fact that $t$-distribution approaches standard normal i.e. we could say the same thing for any other sequence of distributions converging to normal. It could happen that $t$-distribution is a better asymptotic approximation of $\sqrt{n}(\bar X - \mu)/s$ than standard normal and it probably is in most cases. However, both can be used in the asymptotic case.
A: In Ross's book, $\sqrt n (\overline{X} - \mu) / S$ has t distribution requires the nominator to be standard normal random variable (which under CLT satisfies). $\frac{(n-1)S^2}{\sigma^2}$ needs to have chi-square distribution (which satisfies according to this post using Taylor expansion with second order). The final condition will be the independence between $\overline{X}$ and $S^2$ (which I don't know...)
