I need to evaluate the following limit: $$L=\lim_{x \rightarrow 0}\left[\psi\left(1+\frac{a}{2x}\right)-\psi\left(\frac12+\frac{a}{2x}\right)\right]$$
where $\psi(x)$ is the digamma function and $a$ is a positive real number.
I started by writing $$L=\lim_{x \rightarrow +\infty}\left[\psi\left(1+\frac{ax}{2}\right)-\psi\left(\frac12+\frac{ax}{2}\right)\right]$$ and then, using the fact that $\psi(x)\sim\ln x$ for large $x$, I got $L=0$ as the result.
However, this seems wrong to me, and numerical trial suggests this is the case.
Is the result I got correct or wrong? Do you know other techniques to evaluate the limit? Thank you.
Let's denote $$I\left(\frac xa\right)=\psi\left(1+\frac{a}{2x}\right)-\psi\left(\frac12+\frac{a}{2x}\right)$$ It is convenient to use the Gauss' integral representation of $\psi$-function
$$\psi(z)=\int_0^\infty\left(\frac{e^{-t}}t-\frac{e^{-zt}}{1-e^{-t}}\right)dt$$ Then $$I\left(\frac xa\right)=\int_0^\infty\frac{e^{-\frac{at}{2x}}\big(e^\frac t2-1\big)}{e^t-1}dt=2\int_0^\infty\frac{e^{-\frac{at}x}}{e^t+1}dt$$ The integrand is dominated by the integrable function $f(t)=\frac2{e^t+1}$. Therefore, based on the dominated convergence theorem we are allowed to take the limit under the integral sign and find that $$\lim_{x\to0}I\left(\frac xa\right)=2\int_0^\infty\left(\lim_{x\to0}e^{-\frac{at}x}\right)\frac{dt}{e^t+1}=0$$ Using the presentation for Euler polynomials $\frac{2e^{xt}}{e^t+1}=\sum_{n=0}^\infty E_n(x)\frac{t^n}{n!}$ $$\boxed{\,\,I\left(\frac xa\right)\sim\sum_{n=0}^\infty\frac{E_n(0)}{n!}\int_0^\infty e^{-\frac{at}x}t^n=\sum_{n=0}^\infty\left(\frac xa\right)^{n+1}E_n(0)\,\,}$$ Giving that $$E_0(0)=1;\,E_1(0)=-\frac12;\,E_2(0)=0;\,E_3(0)=\frac14;\,E_4(0)=0;\,...$$ $$I\left(\frac xa\right)=\frac xa-\frac12\left(\frac xa\right)^2+\frac14\left(\frac xa\right)^4+O\left(\Big(\frac xa\Big)^6\right)$$