Interesting limit involving gamma function $$\displaystyle  \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}$$
This is from here. 
If we add one more variable - 

$$\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$

Then, the results I got from W|A are really interesting. They seem to follow the pattern which says 

$\displaystyle L(p,q,m,t) = \dfrac{1}{t^{|p-q|/m}}$

If it is indeed true, how to prove it? Factorial approximations did not lead me anywhere.
 A: From Abramowitz and Stegun we know
$$\ln \Gamma(x) = x \ln(x) - \frac{1}{2} \ln(x) - x + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{x}) \qquad (x \to \infty)$$
Applying this formula here yields
$$\begin{align*}
\ln \Gamma\left(\frac{p}{m} + \frac{n}{t}\right) &= \left(\frac{p}{m} + \frac{n}{t}\right) \ln\left(\frac{p}{m} + \frac{n}{t}\right) - \left(\frac{p}{m} + \frac{n}{t}\right) + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{n}) \\
&= \left(\frac{p}{m} + \frac{n}{t}\right) \ln(n) - \frac{n}{t} - \frac{p}{m}
 + \left(\frac{p}{m} + \frac{n}{t}\right) \ln \left(\frac{1}{t} + \frac{p}{mn}\right) + \frac{\ln(2\pi)}{2} + o(1).
\end{align*}$$
After cancelling some terms, we get
$$\ln \frac{\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + \frac{n}{t}\right)} = \frac{p-q}{m} \ln(n) - \frac{p - q}{n} + \frac{p - q}{m} \ln\left(\frac{1}{t}\right) + \frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) + o(1).$$
Now note that
$$\frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) = \frac{1}{t} \ln\left(\frac{\left(1 + \frac{pt}{mn}\right)^n}{\left(1 + \frac{qt}{mn}\right)^n}\right) = \frac{1}{t}\ln\left(\frac{\exp(pt/m)}{\exp(qt/m)}\right) + o(1) = \frac{p - q}{m} + o(1).$$
After cancelling some more terms we get
$$\ln \frac{\Gamma\left(\frac{p}{m} + n\right) \Gamma\left(\frac{q}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + n\right)\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)} = \frac{q - p}{m} \ln(\tfrac{1}{t}) + o(1) = \frac{p - q}{m} \ln(t) + o(1).$$
Using the formula $\Gamma(x + 1) = x \Gamma(x)$ we see that the limit in the last equation is equal to $\ln L(p, q, m, t)$, which finally shows
$$L(p, q, m, t) = t^{(p-q)/m}.$$
A: In the same spirit as Dominik's answer, consider $$A=\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$ and take logarithms.
Now use Stirling approximation $$\log\left( \Gamma\left(x\right)\right)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi
   )-\log(x)\right)+O\left(\frac{1}{x}\right)$$ Apply to each term and simplify as much as you can.
Now, apply to the result Taylor again for large values of $n$ to get (to second order)
$$\log(A)=\frac{(q-p) \log \left(\frac{1}{t}\right)}{m}+\frac{ (q-p) (m+p+q)(t-1)}{2 m^2 n}+O\left(\frac{1}{n^2}\right)\tag 1$$ which makes the limit to be $$\log(A)=\frac{(q-p) \log \left(\frac{1}{t}\right)}{m}=\frac{(p-q) \log \left(t\right)}{m}\implies A= L(p,q,m,t)={t^{\frac{p-q}{m}}}$$ The second term in $(1)$ shows how the limit is approached.
