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Can MMA find limits if the limit can be expressed as a function?

Example:

$$\lim_{x \to \infty}\frac{\Gamma\left(\frac{x+1}{2}\right)}{\Gamma\left(\frac{x}{2}\right)} =\sqrt\frac{x}{2}=\infty$$ $\\\\$

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

returns $\infty$ but I am interested also in the more detailed answer $\sqrt\frac{x}{2}$.

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 $0$.

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    $\begingroup$ Maybe: Series[Gamma[1/2 + x/2]/Gamma[x/2], {x, Infinity, 0}]? $\endgroup$ – Mariusz Iwaniuk Nov 14 '20 at 13:35
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Asymptotic[Gamma[(x + 1)/2]/Gamma[x/2], x -> ∞]

Sqrt[x]/Sqrt[2]

Or

Series[Gamma[(x + 1)/2]/Gamma[x/2], x -> ∞]
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    $\begingroup$ These variants work only in a newer MMA version. $\endgroup$ – granular bastard Nov 14 '20 at 14:44
  • $\begingroup$ @granularbastard : I agree with your comment. I disagree that it is relevant -- the Question gives no M'ma version constraint and is not tagged with any of the various version tags. $\endgroup$ – Eric Towers Nov 14 '20 at 23:27
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Thanks to user Mariusz Iwaniuk the answer can be found easily:

Series[Gamma[1/2 + x/2]/Gamma[x/2], {x, Infinity, 0}] returns

$$\sqrt{\frac{x}{2}}-\frac{1}{4\sqrt{2x}}+\mathcal{O}\left(\frac{1}{x}\right)$$

and the 2 right terms vanish in the limit.

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