Convergence of Hypergeometric Distribution to Binomial Assuming that we have a Hypergeometric distribution:
$$f(x | N, M, K) = \frac{{M \choose x}{N - M \choose K- x}}{{N \choose K}}$$
where $x \in \{0, 1, ..., M\}$ and assume $N, M \rightarrow \infty, \frac{M}{N} \rightarrow p$, then
$f(x | N, M, K) \rightarrow {K \choose x}p^{x}(1 - p)^{K - x}$
In short, as the population $N$ and targets $M$ grow, Hypergeometric converges to Binomial(K, p).
The book suggests solving this by utilising Stirling approximation. I have no question about that. By rewriting the binomial coefficients:
$$\frac{{M \choose x}{N - M \choose K- x}}{{N \choose K}} = \frac{M!(N-M)!K!(N-K)!}{x!(M-x)!(N-M-K+x)!(K-x)!N!} \sim \frac{\sqrt{2\pi}M^{M+1/2}e^{-M}...\sqrt{2\pi}(N-K)^{N-K+1/2}e^{K-N}}{\sqrt{2\pi}x^{x + 1/2}e^{-x}...\sqrt{2\pi}N^{N+1/2}e^{-N}}$$
Then using some algebra we take the limit of this, and arrive at the desired result.
BUT
Is it actually justified? I feel like the author (Casella & Berger, exercise 3.11) intends this to be "the way" to solve this task, hence the "hint" about Stirling formula.
But from googling, Stirling formula is a relative approximation! In short, it is $$\lim\limits_{N \rightarrow \infty}\frac{N!}{\sqrt{2\pi}N^{N + 1/2}e^{-N}} = 1$$
In fact, the factorial in the numerator and Stirling function might not "converge" at all in a traditional sense of this word: there is no $\epsilon$, s.t. $|N! - \text{Stirling}(N)| < \epsilon$ for all $N > \text{some number}$.
Thus, even though I establish the limit for the "Stirling"-representation of the Hypergeometric distribution, I actually cannot establish the transitive relation to the original formula in terms of factorials. Thus, I can find $\epsilon > 0$, s.t. "Stirling representation" is as close to Binomial as I want pointwise, I cannot find any $\epsilon_2$, s.t. Hypergeometric is arbitrarily close to its Stirling-representation pointwise.
Is my logic correct?
If it is, is there a way to show this differently?
Because it would mean that the derivation of convergence is incorrect.
 A: Suppose you have the following relative approximations $\frac{f_1(n)}{g_1(n)} \to 1$ and $\frac{f_2(n)}{g_2(n)} \to 1$.
Then you can use this to help compute the limit of $f_1/f_2$ via
$$\lim_{n \to \infty} \frac{f_1(n)}{f_2(n)}
= \left(\lim_{n \to \infty} \frac{f_1(n)}{f_2(n)}\right)
\frac{\lim_{n \to \infty} g_1(n)/f_1(n)}{\lim_{n \to \infty} g_2(n)/f_2(n)}
= \lim_{n \to \infty}
\frac{f_1(n)}{f_2(n)} \frac{g_1(n)/f_1(n)}{g_2(n)/f_2(n)}
= \lim_{n \to \infty} \frac{g_1(n)}{g_2(n)}.$$
This is what is being hidden in the "$\sim$" step where the factorials are replaced using Stirling's approximation.

Response to comment:
My answer is not talking about $\frac{\text{Hypergeometric PMF}}{\text{Binomial PMF}} \to 1$. You are correct that the desired result is $\text{Hypergeometric PMF} \to \text{Binomial PMF}$.
My answer is talking about why you can replace the factorials in $\frac{M! (N-M)! \cdots}{x! \cdots}$ individually with Stirling's approximation (the "$\sim$" step that your question is asking about).
Specifically, the authors really are doing
\begin{align}
\lim_{M,N \to \infty} \frac{{M \choose x}{N - M \choose K- x}}{{N \choose K}}
&=  \lim_{M,N \to \infty} \frac{M!(N-M)!K!(N-K)!}{x!(M-x)!(N-M-K+x)!(K-x)!N!}
\\
&= \lim_{M,N \to \infty}  \frac{\sqrt{2\pi}M^{M+1/2}e^{-M}...\sqrt{2\pi}(N-K)^{N-K+1/2}e^{K-N}}{\sqrt{2\pi}x^{x + 1/2}e^{-x}...\sqrt{2\pi}N^{N+1/2}e^{-N}}
\end{align}
and the my answer above is justifying the last equality. Dropping the limits and using "$\sim$" is a common shorthand.
A: You ask for a way to show this differently.
I leave as an exercise for the reader the fussy correction for $M = [nP-1/2,nP+1/2) \cap \Bbb{Z}$, that is letting $M$ always be the integer closest to $pN$ as $N$ varies.  (Alternatively, replace all the factorials with Gammas and let $M$ and $N$ range smoothly through real numbers instead of through integers.)  Here, we proceed by approximating $M$ with $nP$ immediately.
You want
\begin{align*}
L &= \lim_{N \rightarrow \infty} \frac{(pN)!(N-pN)!K!(N-K)!}{x! (pN - x)! (K-x)!(N-pN-K+x)!N! }  \\
&= \frac{K!}{x! (K-x)!} \cdot \lim_{N \rightarrow \infty} \frac{(N-K)!}{K!} \cdot \frac{(pN)!}{(pN-x)!}  \\
    &\qquad \cdot \frac{(N-pN)!}{(N-pN-K+x)!}  \text{.}
\end{align*}
This really is just an application of "factor out the big part".  There are three cases.  I show the $x > K$ case.  The $x = K$ case is similar, but a little easier since we have $\frac{1}{N^0(1-p)^0}$ in the second step.  The $x < K$ case is analogous to the one shown.  So, suppose $x > K$ ...
\begin{align*}
L &= \binom{K}{x} \lim_{N \rightarrow \infty} \frac{1}{N(N-1)(N-2) \cdots (N-K+1)}  \\
&\qquad \cdot \frac{(pN)(pN-1) \cdots (pN-x+1)}{1}  \\
&\qquad \cdot \frac{1}{(N-pN-K+x) \cdots (N-pN+1)}  \\
    &= \binom{K}{x} \lim_{N \rightarrow \infty} \frac{1}{N^K}\frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)}  \\
&\qquad \cdot p^xN^x \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1}  \\
&\qquad \cdot \frac{1}{N^{x-K}(1-p)^{x-K}}\frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)}  \\
    &= \binom{K}{x} p^x \frac{1}{(1-p)^{x-K}}  \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)}  \\
&\qquad \cdot \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1}  \\
&\qquad \cdot \frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)}  \\
    &= \binom{K}{x} p^x (1-p)^{K-x}  \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{1 \left( 1-\frac{1}{N} \right) \left( 1-\frac{2}{N} \right) \cdots \left( 1- \frac{K-1}{N} \right)}  \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{(1) \left( 1- \frac{1}{pN} \right) \cdots \left( 1-\frac{x-1}{pN} \right)}{1}  \\
&\qquad \cdot \lim_{N \rightarrow \infty} \frac{1}{\left(1+ \frac{-K+x}{N-pN} \right) \cdots \left( 1+ \frac{1}{N-pN} \right)}  \\
    &= \binom{K}{x} p^x (1-p)^{K-x} \cdot 1 \cdot 1 \cdot 1  \text{.}
\end{align*}
(For evaluating the limits involving products of terms of the form $\displaystyle \left( 1 \pm \frac{A(K,p,x)}{N} \right)$, multiply them out completely, obtaining $1 \pm ($ terms involving $K,p,x$ and various negative powers of $N)$.  Of course as $N \rightarrow \infty$, the terms involving negative powers of $N$ go to $0$, leaving only the leading $1$.
