Evaluate $\lim_{n\to 0}{\ln\Gamma(2+n)\over n(n+1)}$ How to does one show that 
$$\lim_{n\to 0}{\ln\Gamma(2+n)\over n(n+1)}=1-\gamma?\tag1$$

$$\lim_{n\to 0}{\ln\Gamma(2+n)\over n}-\lim_{n\to 0}{\ln\Gamma(2+n)\over n+1}$$

We have this well-known

$$\gamma=\lim_{n\to \infty}\left(n-\Gamma\left({1\over n}\right)\right)$$
Using the digamma function defined by
$$\psi(x) = \frac{d}{dx} \ln(\Gamma(x)) = \frac{\Gamma'(x)}{\Gamma(x)}$$
where $\psi(1) = - \gamma$ and $\psi(x+1) = \psi(x) + \frac{1}{x}$
then
\begin{align}
\lim_{n \to 0} \, \frac{\ln\Gamma(n+2)}{n (n+1)} &amp;= \frac{1}{0+1} \, \lim_{n \to 0} \frac{\ln\Gamma(n+2)}{n} \\
&amp;= \lim_{n \to 0} \frac{\Gamma'(n+2)}{\Gamma(n+2)} \cdot \frac{1}{1} \\
&amp;= \lim_{n \to 0 } \psi(n+2) = \psi(2) = \psi(1) + \frac{1}{1} \\
&amp;= 1 - \gamma.
\end{align}
Using the digamma function defined by
$$\psi(x) = \frac{d}{dx} \ln(\Gamma(x)) = \frac{\Gamma'(x)}{\Gamma(x)}$$
where $\psi(1) = - \gamma$ and $\psi(x+1) = \psi(x) + \frac{1}{x}$
then
\begin{align}
\lim_{n \to 0} \, \frac{\ln\Gamma(n+2)}{n (n+1)} &amp;= \frac{1}{0+1} \, \lim_{n \to 0} \frac{\ln\Gamma(n+2)}{n} \\
&amp;= \lim_{n \to 0} \frac{\Gamma'(n+2)}{\Gamma(n+2)} \cdot \frac{1}{1} \\
&amp;= \lim_{n \to 0 } \psi(n+2) = \psi(2) = \psi(1) + \frac{1}{1} \\
&amp;= 1 - \gamma.
\end{align}