How do I calculate this limit when two terms tend to infinity at similar rates In a particular problem that I am currently trying to solve, I have the following expression (this is not the entire expression, I have included only the terms involving $a_1$ and $b_1$),

$\lim_{(a_1,b_1)\to (\infty,\infty)} \frac{(a_1+a_2+p)!}{(a_1 + p)!}\frac{(b_1+q)^{a_1 +p}}{(b_1+b_2+q)^{a_1+a_2+p}}$ [equation 1]

$a_1$ and $b_1$ tend to infinity such that $a_1/b_1 \rightarrow c_0$, i.e., the ratio of $a_1$ and $b_1$ tends to a fixed value. (It is known that $a_2$, $p$ and $q$ do not tend to infinity)

This is what I did so far. I used Stirling's approximation for the factorials, the limit reduces to,
$\lim_{(a_1,b_1)\to (\infty,\infty)} \left[\sqrt{\frac{a_1+a_2+p}{a_1 +p}}\left(\frac{a_1+a_2+p}{b_1+b_2+q}\right)^{a_1+a_2+p} \left(\frac{b_1+q}{a_1+p}\right)^{a_1+p} e^{-a_2}\right]$
I then, made these approximations, $(a_1+a_2+p) \rightarrow a_1$, $(b_1+b_2+q) \rightarrow b_1$, $(a_1+p) \rightarrow a_1$ and $(b_1+q) \rightarrow b_1$. With these approximations, the limit is calculated as,
$\lim_{(a_1,b_1)\to (\infty,\infty)} \left(\frac{a_1}{b_1}\right)^{a_2}e^{-a_2} = c_0^{a_2}e^{-a_2}$ [equation 2]

I want to know if this is a correct way to approximating the limit. I doubt my solution because the simulation results in values that can be obtained if I approximate [equation 1] as $c_0^{a_2}e^{-c_0}$. This differs from [equation 2] in only one term but creates a huge difference. Is there any problem in my the way I approximate  the limit?
How do I solve this? Any help is appreciated. Thanks in advance
 A: Yes, there is a problem in your approximations. Recall that
$$\lim_{n \to \infty} \biggl(1 + \frac{x}{n}\biggr)^n = e^x.$$
Effectively, you have replaced this limit with $1$ four times, and since the various $x$s don't cancel, you got a wrong result.
The square root does indeed tend to $1$, so we can ignore that. Assuming that $a_2$ and $p$ are constant and $b_2$ and $q$ remain bounded, we get
$$\biggl( \frac{a_1 + a_2 + p}{b_1 + b_2 + q}\biggr)^{a_2 + p} \to c_0^{a_2 + p}\quad\text{and}\quad \biggl(\frac{b_1 + q}{a_1+p}\biggr)^p \to c_0^{-p},$$
so together we get a factor of $c_0^{a_2}$ from these two.
We can regroup the remaining factors with the common exponent $a_1$:
\begin{align}
\biggl(\frac{a_1 + a_2 + p}{b_1 + b_2 +q}\biggr)^{a_1}\biggl(\frac{b_1+q}{a_1+p}\biggr)^{a_1} &= \biggl(\frac{a_1 + a_2 + p}{a_1+p}\biggr)^p\biggl(\frac{b_1 + b_2+q}{b_1+q}\biggr)^{-a_1}\\
&= \biggl( 1 + \frac{a_2}{a_1+p}\biggr)^{a_1}\biggl( 1 + \frac{b_2}{b_1+q}\biggr)^{-a_1}.
\end{align}
Now $\frac{a_1+p}{a_1} \to 1$, and so
$$\biggl( 1 + \frac{a_2}{a_1+p}\biggr)^{a_1} \to e^{a_2}.$$
And $\frac{a_1}{b_1+q} \to c_0$, so if $b_2$ is constant, we have
$$\biggl(1 + \frac{b_2}{b_1+q}\biggr)^{a_1} \to e^{c_0\cdot b_2}.$$
Thus altogether, under the stated assumptions we find (if I haven't miscalculated)
$$\lim_{\substack{(a_1,b_1)\to (\infty,\infty) \\ a_1/b_1 \to c_0}} \frac{(a_1+a_2+p)!}{(a_1+p)!}\frac{(b_1+q)^{a_1+p}}{(b_1+b_2+q)^{a_1+a_2+p}} = c_0^{a_2}\cdot e^{-b_2c_0}.$$
We can relax the constancy assumptions, it suffices that $p$ and $q$ remain bounded, and that $a_2$ and $b_2$ converge to obtain the result.
In your simulation, it seems that $b_2 = 1$ resp. $b_2 \to 1$.
