Evaluate the limit $\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$ Evaluate
$$\lim_{n\to\infty}\left[n+{n^2}\log{\frac{n}{n+1}}\right]$$
I know that this limit is equal to $\frac1{2}$ but I don't know how to do it.
Thanks!
 A: Let $n=1/x
$$$\lim_{n\to\infty}\left[n-{n^2}\log{\frac{n+1}{n}}\right]=\lim_{x\to0}\left[\frac{x-\log(1+x)}{x^2}\right]=\lim_{x\to0} \frac{1}{2(x+1)}=\frac{1}{2}$$
where the penultimate equality is obtained by  l'Hôpital's rule.
Note: as Marvis well noticed, the first equality is true because $\lim_{x\to0}\left[\frac{x-\log(1+x)}{x^2}\right]$ exists. 
Chris.
A: Recall that $\log(1-x) = -x - \dfrac{x^2}2 + \mathcal{O}(x^3)$. Hence,
\begin{align}
\log \left(\dfrac{n}{n+1}\right) & = \log \left(1-\dfrac1{n+1}\right)\\
 & = -\dfrac1{n+1} - \dfrac1{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)^3}\right)\\
n^2\log \left(\dfrac{n}{n+1}\right) & = -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2}+ \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
n + n^2\log \left(\dfrac{n}{n+1}\right) & = n -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
& = \dfrac{n^2+n-n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
& = \dfrac{n}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\
\end{align}
Hence, the limit is $\dfrac12$.
A: Since $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$ and $\log\left(\frac{n}{n+1}\right)=-\log\left(1+\frac1n\right)$,
$n+n^2\log\left(\frac{n}{n+1}\right)=n-n^2\log\left(1+\frac1n\right)=n-n^2\left(\frac1n-\frac12\frac1{n^2}+o\left(\frac1{n^2}\right)\right)=\frac12+o(1).$
