Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to
\infty} s_n$. 
 \begin{align}
  s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx
 \end{align}
I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions. 

Here's a solution based on order statistics, similar to my answer here.
Let $X_1,\dots, X_n$ be i.i.d. uniform(0,1) random variables. 
The distribution function of $X$ is $F(x)=x$ and density $f(x)=1$ for $0\leq x\leq 1$.
Now let $M=\max(X_1,\dots, X_n)$; its density function is 
$$f_M(x)=n F(x)^{n-1}f_X(x)=n\,x^{n-1}\text{ for }0\leq x\leq 1.$$ 
Also, it is not hard to see that  $M\to 1$ in distribution as $n\to\infty$.
Now $$\int_0^1 {n x^{n-1}\over 1+x} \,dx =\int_0^1 {1\over 1+x}\, f_M(x) \,dx
=\mathbb{E}\left({1\over 1+M}\right).$$ 
This converges to 
${1\over 1+1}={1\over 2}$ as $n\to\infty$.
