Limit if the limit is a function

Question

Can MMA find limits if the limit can be expressed as a function?
Example:
$$lim_{x to infty}frac{Gammaleft(frac{x+1}{2}right)}{Gammaleft(frac{x}{2}right)} =sqrtfrac{x}{2}=infty$$
$\$
Limit[Gamma[(x + 1)/2] / Gamma[x/2], x -> ∞]
returns $infty$  but I am interested also in the more detailed answer $sqrtfrac{x}{2}$.
So far only in case I presume the answer I could check if it's true:
Limit[Gamma[(x + 1)/2] / Gamma[x/2] - Sqrt[x/2], x -> ∞] returns $0$.

granular bastard · Answer

Thanks to user Mariusz Iwaniuk the answer can be found easily:
Series[Gamma[1/2 + x/2]/Gamma[x/2], {x, Infinity, 0}] returns
$$sqrt{frac{x}{2}}-frac{1}{4sqrt{2x}}+mathcal{O}left(frac{1}{x}right)$$
and the 2 right terms vanish in the limit.

cvgmt · Answer

Asymptotic[Gamma[(x + 1)/2]/Gamma[x/2], x -> ∞]

Sqrt[x]/Sqrt[2]

Or
Series[Gamma[(x + 1)/2]/Gamma[x/2], x -> ∞]

Limit if the limit is a function

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