Difficult Limit involving digamma function Evaluate: $$\lim_{z \to 0} \psi(-z)\cdot \bigg ( 1 - 2z(z+1) \bigg) - z\cdot\psi'(-z) $$ If we simply substitute in $0$ that gets us infinity, and problems. The answer is $-2 - \gamma$ How do we solve limits with digamma? Should I use series? Please help! Thanks! **Full answer is not needed, just a starting point!**

Since we can write this limit as a residue, an answer is already given here.

Anyway, since in a punctured neighbourhood of the origin we have: $$ \psi(-x)=\frac{1}{x}-\gamma+\zeta(2)x+\ldots, $$ $$-\psi'(-x) = -\frac{1}{x^2}+\zeta(2)+\ldots,$$ it follows that: $$ (1-2x(1+x))\,\psi(-x) = -\frac{1}{x}-(2+\gamma)+\left(2+2\gamma+\zeta(2)\right) x+\ldots,$$ so the limit is $-(2+\gamma)$ as wanted.