Unlock dozens of cosmetic enhancements and cool Mega Mic abilities as you level up. Battle it out in six versus modes and chase your opponents to the top of the leaderboard. If you miss your golden opportunity to gain an upgrade, never fear. The loot slider is there for you. Mega Machines comes to the mobile phone of your dreams. Don’t wait. Get this game now!Q:

Tricky limit involving factorials and polygamma function

The following appears as a lemma in a paper by Chafaï (2013), p. 798, Appendix B.
Let $x,\mu\in(0,\pi/2)$, and
$$
z=\frac{x}{2}-\frac{\mu}{x}.
$$
Further,
$$
\frac{1}{\pi}\mathrm{Re}\psi(1+iz)
=\frac{1}{2\pi}\log\left|\frac{1}{1+iz}\right|
-\frac{\Gamma'(1+iz)}{2\pi(1+iz)},
$$
where $\psi(z)$ is the polygamma function.
Then,
\begin{align*}
\lim_{x\downarrow 0}\left(\frac{2}{\pi x}\mathrm{Re}\psi(1+iz)\right)^{\frac{1}{1+iz}}
&=\frac{2}{\pi}\mathrm{Re}\psi'(1+iz)\\
&=\frac{1}{2\pi}\log\left|\frac{1+iz}{iz}\right|-\frac{1}{2\pi}.
\end{align*}
The proof is done in such a way that it requires only some elementary properties of the polygamma function and it holds for every $x\in(0,\pi/2)$, i.e. also for $x=0$.

I suspect that this limit is well-known and that the reasoning is quite elementary. I also suspect that the substitution $z=x/(2z)$ is “unnecessary”.

A:

Let $\mu := \pi/x$. Then, as $x \to 0$ we have $\mu \to 0$.
Then, as $x \to 0$ we have

 

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