$\displaystyle \lim_{n\to\infty}n-n^{2}log\frac{n+1}{n}$
Sorry for occupying the space, but the idea came to me when solving other limit.
Maybe it will help someone else.
The initial limit can be written as:
$\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n}-log(1+\frac{1}{n})}{\frac{1}{n^{2}}}$
We can substitute 1/n with x, x real, and we get:
$\displaystyle \lim_{x\to0}\frac{x-log(1+x)}{x^{2}}$
After applying 2 times l'Hopital's rule we get the final answer 1/2.
I believe there is another solution without l'Hopital and functions, only with sequences .
Note that for arbitrarily large $\displaystyle n$ it is true that $\displaystyle \ln\left(\frac{n+1}{n}\right)=\ln\left(1+\frac{1}{ n}\right)\underset{n\to\infty}{\sim}\frac{1}{n}-\frac{1}{2n^2}$ so then $\displaystyle \lim_{n\to\infty}\left\{n-n^2\ln\left(\frac{n+1}{n}\right)\right\}=\lim_{n\t o\infty}\left\{n-n^2\left(\frac{1}{n}-\frac{1}{2n^2}\right)\right\}=\frac{1}{2}$.