Factorial limits $\lim _{x\to \infty }\frac{\left(x!\right)^3\left(3\left(x+1\right)\right)!}{\left(\left(x+1\right)!\right)^3\left(3x\right)!}$ $$\lim _{x\to \infty }\frac{\left(x!\right)^3\left(3\left(x+1\right)\right)!}{\left(\left(x+1\right)!\right)^3\left(3x\right)!}$$
i read this Calculating limit involving factorials.
I don't understand how they eliminate !
The answer is 27.
 A: Hint:
$$(x!/(x+1)!)^3=1/(x+1)^3$$ and  $$(3(x+1)!)/(3x)!=(x+1)(x+2)(x+3)$$
A: Hint : try to simplify first terms of numerator and denominator and then second terms and then take $x^3$ common and then evaluate the limits.
A: $$\lim _{x\to \infty }\frac{\left(x!\right)^3\left(3\left(x+1\right)\right)!}{\left(\left(x+1\right)!\right)^3\left(3x\right)!} = \lim _{x\to \infty }\left( \frac{x!}{\left(x+1\right)!}\right)^3\cdot \frac {(3x+3))!}{(3x)!} \\= \lim _{x\to \infty }\frac{(3x+1)(3x+2)(3x+3)}{(x+1)^3} = 27.$$
A: If you use Stirling approximation $$n! \approx \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ after some basic simplifications, you will find that $$A=\frac{\left(x!\right)^3\left(3\left(x+1\right)\right)!}{\left(\left(x+1\right)!\right)^3\left(3x\right)!}\approx \frac{27 x}{x+1}$$ If you use Starling $$n! \approx \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}e^{\frac{1}{12n}}$$ you will find that $$A \approx \frac{27 x}{x+1}e^{\frac{2}{9 x^2+9 x}}$$ There re other useful approximations for $n!$ (Gosper, Burnside,...) but, for the limiting behavior, all of them lead to the same result.
A: Hints:
$$\lim _{x\to \infty }\frac{\left(x!\right)^3\left(3\left(x+1\right)\right)!}{\left(\left(x+1\right)!\right)^3\left(3x\right)!}=\lim _{x\to \infty }\frac{(3x+3)(3x+2)(3x+1)}{\left(x+1\right)^3}=27.$$
