Thinking about a recent question of Claude Leibovici I come across the limit :
$$\lim_{x\to \pi^{-}}\Big(\tan\Big(x+\frac{\pi}{2}\Big)-\Gamma(\sin(x))\Big)=\gamma$$
Proof of the limit above :
It's equivalent to :
$$\lim_{x\to \pi^{-}}\Big(-\frac{\cos(x)}{\sin(x)}-\Gamma(\sin(x))\Big)=\gamma$$
With the substitution $y=\sin(x)$ we got :
$$\lim_{y\to 0^{+}}\Big(-\frac{\sqrt{1-y^2}}{y}-\Gamma(y)\Big)=\gamma$$
Now we use the fact :
$$\Gamma(y)=\frac{1}{y}-\gamma+(6\gamma^2+\pi^2)y+O(y^2)$$
The limit becomes :
$$\lim_{y\to 0^{+}}\Big(-\frac{\sqrt{1-y^2}}{y}-(\frac{1}{y}-\gamma)\Big)=\gamma$$
Wich is not hard to solve.
Now I would like to prove :
$$\Big(\tan\Big(x+\frac{\pi}{2}\Big)-\Gamma(\sin(x))\Big)<C+(\pi-x)B \quad (1)$$
for $3<x<\pi$ and the $C,B$ are positive constants.
I think we can use the limit describes aboves .
Question :
Is there a nice way to prove the inequality $(1)$ and find $B$ ?
Thanks in advance .
Regards Max
You can do it composing Taylor expansion and get for $$y=\tan\Big(x+\frac{\pi}{2}\Big)-\Gamma(\sin(x))$$ $$y=\gamma -\frac{6+6 \gamma ^2+\pi ^2}{12} (\pi-x )+\frac{4 \zeta (3)+2 \gamma ^3+\gamma \pi ^2}{12} (\pi-x )^2 +O\left((\pi-x )^3\right)$$