Let:
- $f(x) = \left(\Gamma(x+1)\right)^{\left(\frac{\ln x}{x\ln x - 1.25506x}\right)} + x - 1$
- $g(x) = \dfrac{f(x)}{x}$
Is $g(x)$ a decreasing function toward a limit? If yes, what is the limit?
I suspect that it is. When I test out values, I am seeing that for $x \ge 17$, $g(x) < 3$ and from there, it appears to decrease toward a limit.
Here is my reasoning about the equation (please feel free to point out any mistakes, I am learning and am glad to revise):
(1) $\dfrac{\ln x}{x \ln x - 1.25506x}$ is decreasing for $x \ge 10$ since in this case $\ln x > 2.25506$ and $x\ln x - 1.25506x > x > \ln x$
(2) Using Stirling's Formula:
$$\Gamma(x+1) \sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$$
(3) $\Gamma(x+1)^{\left(\frac{\ln x}{x\ln x - 1.25506x}\right)} \sim \left(\sqrt{2\pi x}\left(\frac{x}{e}\right)^x\right)^{\frac{\ln x}{x}}$
(4) Would it be correct to assume that this approximates to:
$$\left(\sqrt{2\pi x}\left(\frac{x}{e}\right)\right)^{\ln x}$$
(5) Since this value appears to grow slower than $x$, the function is decreasing.
(6) To confirm that it is decreasing, I would then take the derivative of $g(x)$ and show that it is less than $0$.
(7) I am not clear how to establish that it has a limit or estimate what that limit is.
Any tips or suggestions for correcting this argument and showing that there is a limit or there is not a limit would be very helpful.