The Taylor expansion of $\Gamma(z+1)$ at the origin is
$$ \Gamma(z+1) = \Gamma(1) + \Gamma'(1) z + \mathcal{O}(z^{2}) = 1- \gamma z + \mathcal{O}(z^{2}) $$
$$ \implies \Gamma(z) = \frac{1}{z} - \gamma + \mathcal{O}(z) $$Therefore,
$$ \lim_{z \to 0} \left( \Gamma(z) -\frac{1}{z} \right) = \lim_{z \to 0} \Big( - \gamma + \mathcal{O}(z) \Big) = - \gamma $$
For all complex values $z$, the Riemann zeta function has the integral representation $$ \zeta(s) = 2 \int_{0}^{\infty} \frac{\sin (z \arctan t)}{(1+t^{2})^{z/2} (e^{2 \pi t} - 1)} \ dt + \frac{1}{2} + \frac{1}{z-1} $$
http://mathhelpboards.com/challenge-questions-puzzles-28/another-integral-representation-riemann-zeta-function-6398.htmlTherefore,
$$ \lim_{z \to 1} \Big( \zeta(s) - \frac{1}{z-1} \Big) = 2 \int_{0}^{\infty} \frac{t}{t^2+1} \frac{1}{e^{2 \pi t}-1}\ dt + \frac{1}{2}$$Differentiating Binet's log-gamma formula,
Binet's Log Gamma Formulas -- from Wolfram MathWorld $$-2 \int_{0}^{\infty} \frac{t}{t^2+z^{2}} \frac{1}{e^{2 \pi t}-1}\ dt = \psi(z) -\log z -1 + \frac{1}{2z} + 1 $$
$$ \implies 2 \int_{0}^{\infty} \frac{t}{t^2+1} \frac{1}{e^{2 \pi t}-1}\ dt = -\psi(1) - \frac{1}{2} = \gamma - \frac{1}{2}$$So
$$ \lim_{z \to 1} \Big( \zeta(s) - \frac{1}{z-1} \Big)= \gamma$$