TITLE: Show that $\lim_{n\to\infty}n+n^2 \log\left(\frac{n}{n+1}\right)= 1/2$
QUESTION [5 upvotes]: Show that  $$\lim_{n\to\infty}\left[n+n^2 \log\left(\frac{n}{n+1}\right)\right]= \frac{1}{2}$$
I can't get how this limit comes about. It says that it comes from the sequential criterion of limits.

REPLY [2 votes]: Let $n=\frac{1}{t}\implies t\to 0$ as $n\to \infty$
$$\lim_{n\to \infty}\left(n+n^2\log\left(\frac{n}{n+1}\right)\right)$$
$$=\lim_{t\to 0}\left(\frac{1}{t}+\frac{1}{t^2}\log\left(\frac{1}{1+t}\right)\right)$$
$$=\lim_{t\to 0}\left(\frac{\underbrace{t-\log\left(1+t\right)}_{\longrightarrow0}}{\underbrace{t^2}_{\longrightarrow0}}\right)$$
using L'Hospital's rule for $\frac 00$ form, 
$$=\lim_{t\to 0}\left(\frac{1-\frac{1}{1+t}}{2t}\right)$$
$$=\lim_{t\to 0}\left(\frac{t}{2t(t+1)}\right)$$
$$=\frac 12\lim_{t\to 0}\left(\frac{1}{t+1}\right)$$
$$=\frac 12\left(\frac{1}{0+1}\right)=\color{red}{\frac 12}$$