\lim_{n\to \infty } \frac{\ln (2 \pi )+\ln (n)}{n+1}
\begin{array}{c}
 \frac{\infty }{\infty } \\
 
\begin{array}{c}
 = \\
 L.P.
\end{array}

\end{array}
\lim_{n\to \infty } \frac{\frac{1}{n}}{1}=0