f’(x)=+0or-0?

$$\begin{array}{rl}{\displaystyle f^{\prime }(x)}& =\log (\frac{x}{x+1})+\frac{1}{x+1}\\ & =\log (1-\frac{1}{x+1})+\frac{1}{x+1}\end{array}$$
${\displaystyle \frac{1}{x+1}=t}$とおけば

$f^{\prime }(x)=\log (1-t)+t(=g(t)\text{とおく})$

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{1}{x+1}=+0}$より$x\to \mathrm{\infty }$で$t\to +0$

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=\underset{t\to +0}{lim}g(t)}$ であり

この公式を用いれば
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)}& =\underset{t\to +0}{lim}g(t)\\ & =\underset{t\to +0}{lim}-t-\frac{t^2}{2}-\frac{t^3}{3}-\cdots +t\\ & =\underset{t\to +0}{lim}-(\frac{t^2}{2}+\frac{t^3}{3}+\cdots )\\ & =-0\end{array}$$

よって${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=-0}$