# Limit if the limit is a function

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 $$0$$.

• Maybe: Series[Gamma[1/2 + x/2]/Gamma[x/2], {x, Infinity, 0}]? – Mariusz Iwaniuk Nov 14 '20 at 13:35

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


Sqrt[x]/Sqrt[2]

Or

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

• These variants work only in a newer MMA version. – granular bastard Nov 14 '20 at 14:44
• @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. – Eric Towers Nov 14 '20 at 23:27

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.