TITLE: Finding the limit of a sequence of integrals
QUESTION [7 upvotes]: Let us define a sequence of function as
  $$f_n(x)=\frac{2nx^{n-1}}{x+1}\;\;\text{for each $x\in [0,1]$ and for all $n\in\mathbb{N}$}$$
  What is $\displaystyle \lim_{n\to \infty} \int_0^1 f_ n(x) dx$ ?

How to find the limit? If I can interchange the limit with integral the ans is surely $0$. But can we interchange the limit with integral here. All that I know is Lebesgue's monotone convergence theorem and dominated convergence theorem that allow us to interchange limit and integrals. But this seems not to be useful here. Then how to proceed? Any help would be appreciated. Thanks in advance.

REPLY [3 votes]: The $nx^{n-1}$ factor suggests we could try using the $y=x^n$ substitution. This way we get
\begin{equation}
\int_0^1\frac{2}{x+1}nx^{n-1}dx = \int_0^1\frac{2}{x+1}(x^n)'dx = \int_0^1\frac{2}{\sqrt[n]{y}+1}dy.
\end{equation}
Now, as $\frac{2}{\sqrt[n]{y}+1}\leq2$, we can actually use Lebesgue's dominated convergence theorem to compute the limit. We know that $\sqrt[n]{y}\to1$ a.e. in $[0,1]$, so
\begin{equation}
\lim_{n\to\infty}\int_0^1\frac{2}{\sqrt[n]{y}+1}dy=\int_0^1\frac{2}{1+1}dy=\int_0^11\,dy=1.
\end{equation}