Factorial limit from gamma function calculation I want to show that $$\lim_{n\rightarrow\infty}\dfrac{\Gamma\left(n+\frac12\right)}{\sqrt{n}\Gamma(n)}=1$$
Using the formula for $\Gamma\left(n+\frac12\right)$ here, it reduces to $$\lim_{n\rightarrow\infty}\dfrac{(2n)!\sqrt{\pi}}{2^{2n}\sqrt{n}(n-1)!n!}=1$$
How to prove that?
 A: See "Inequalities for Gamma Function Ratios", 
G.J.O. Jameson, American Mathematical Monthly, 
December 2013, pp 936-940.
This recent article gives elementary proofs of 
Gautschi-type inequalities. In particular, you 
get 
$$\left({x\over x+1/2}\right)^{1/2}\leq {\Gamma(x+1/2)\over\sqrt{x}\,\Gamma(x)}\leq 1.$$
Now let $x\to\infty$.
A: Notice that the reciprocal of the expression under the limit is just the Euler's beta function (see wiki for details):
$$
   \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} = \operatorname{B}\left(n, \frac{1}{2}\right) = \int_0^1 \frac{x^{n-1}}{\sqrt{1-x}} \mathrm{d}x
$$
Hence, using $\Gamma(1/2)=\sqrt{\pi}$:
$$
   \lim_{n \to +\infty} \frac{\sqrt{n} \Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}  = \lim_{n \to +\infty} \int_0^1 \frac{\sqrt{n}}{\sqrt{\pi}} \frac{x^{n-1}}{\sqrt{1-x}} \mathrm{d}x \stackrel{x=1-t/n}{=} \lim_{n \to +\infty} \int_0^n \frac{1}{\sqrt{\pi}} \frac{\left( 1-\frac{t}{n}\right)^{n-1}}{\sqrt{t}} \mathrm{d}t 
$$
Using dominated convergence theorem:
$$
  \lim_{n \to +\infty} \int_0^n \frac{1}{\sqrt{\pi}} \frac{\left( 1-\frac{t}{n}\right)^{n-1}}{\sqrt{t}} \mathrm{d}t  = \frac{1}{\sqrt{\pi}} \int_0^\infty t^{-1/2} \underbrace{\lim_{n \to \infty} \left(1-\frac{t}{n}\right)^{n-1}}_{\exp(-t)} \mathrm{d}t
= \frac{1}{\sqrt{\pi}} \underbrace{\int_0^\infty t^{-1/2} \mathrm{e}^{-t} \mathrm{d}t}_{\Gamma\left(1/2\right)=\sqrt{\pi}} = 1
$$
Since we established that
$$
\lim_{n \to +\infty} \frac{\sqrt{n} \Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}  = 1   
$$
the limit in question follows.
A: You are going around in circles, since you reduced a problem with two gammas to a problem with three gammas! Instead, the scientific way uses Stirling's expansion. Notice that the second (an in some ways easier) derivation is used for Gamma directly. (in any case, use the Stirling approximation for numerator and denominator, divide, take limits.
