A calculus problem by Kenny Lau

the expression can be written as $\displaystyle \lim_{x\to 0} \frac{1-x\Gamma(x)}{x}=\lim_{x\to 0}\frac{1-\Gamma(x+1)}{x}$

Hence by L'Hopital rule

$\displaystyle \lim_{x\to 0}\frac{1-\Gamma(x+1)}{x}=-\Gamma'(1)=-\Gamma(1)\psi(1)=-\psi(1)=\gamma=0.5772$

Here $\psi(.)$ denotes the Digamma Function and $\gamma$ denotes the Euler Mascheroni Constant .