Steps in CLT proof unclear

Question

In john rice Mathematical Statistics and Data Analysis we find a proof about the central limit theorem.

Let $X_1, X_2,\ldots$ be a sequence of independent random variables having variance $\sigma^2$ and the commmon distribution function $F$ and moment generating function $M$ defined in a neighborhood of zero. Let

$$S_n = \sum_{i = 1}^n X_i$$

Then:

$$\lim_{n \longrightarrow \infty}P \left( \frac{S_n}{\sigma \sqrt n} \le x\right) = \Phi(x)$$

Proof:

Let $Z_n = \frac{S_n}{\sigma \sqrt{n}}$

Since $S_n$ is a sum of independent random variables:

$$M_{S_n}(t) = [M(t)]^n$$

$$M_{Z_n}(t) = \left[M\left(\frac t {\sigma \sqrt n}\right)\right]^n$$

$M(s)$ has a Taylor series expansion about zero.

$$M(s) = M(0) + s M'(0) + \frac{1}{2}s^2 M''(0) + \varepsilon_s$$

where $\frac{\varepsilon_s}{s^2} \longrightarrow 0$ as $s \longrightarrow 0$.

Why is this? And why is it useful? I thought that if the denominator of a fraction converges to zero you would essentially have an expression that have a zero value in the denominator which would produce an undefined value.

Since $E(X) = 0$, $M'(0) = 0$ and $M''(0) = \sigma^2$.

As $n \longrightarrow \infty, t/(\sigma \sqrt{n}) \longrightarrow 0$

The statement $n \longrightarrow \infty, t/(\sigma \sqrt{n}) \longrightarrow 0$ made, but why is it useful?

$$M\left(\frac{t}{\sigma \sqrt{n}}\right) = 1 + \frac{1}{2}\sigma^2 \left(\frac{t}{\sigma \sqrt{n}}\right)^2 + \varepsilon_n$$

I assume the above follows from the fact that $s = \frac{t}{\sigma \sqrt{n}}$ So you replace this into the taylor expansion, but why is $\varepsilon_{s}$ replaced with $\varepsilon_n$?

Answer 1

if the denominator of a fraction converges to zero you would essentially have an expression that have a zero value in the denominator which would produce an undefined value."

THAT IS FALSE. All derivatives are limits as the numerator and denominator approach $0.$ If that made them undefined, then all derivatives would be undefined. If the denominator approaches $0$ while the numerator approaches something else, then the limit is undefined, but if they BOTH approach $0,$ then in many cases the limit exists.

Note that if the numerator is $s^3$ and the denomoinator is $s^2,$ then certainly the fraction approaches $0$ as $s^2\to0.$