Limit and definite integral question I recently came across this problem:

What is the value $$\mathrm{\lim_{n\to\infty}\int_0^1 \frac{nx^{n-1}}{1+x}dx}$$ Assume that the interchange of limit and integration is possible.

I have no idea how to begin solving this problem. I guess expressing the integral as an inequality might help, but I could not make any way out by that method. Any ideas?
 A: Integrating by parts we get
$$\int_0^1\frac{nx^{n-1}}{1+x}dx=\left[\frac{x^n}{1+x}\right]_0^1+
\int_0^1\frac{x^{n}}{(1+x)^2}dx$$
So
$$0\le\int_0^1\frac{nx^{n-1}}{1+x}dx-\frac{1}{2}=
\int_0^1\frac{x^{n}}{(1+x)^2}dx\le \int_0^1{x^{n}}dx=\frac{1}{n+1}$$
Taking the limit as $n$ tend to $+\infty$ we get
$$\lim_{n\to\infty}\int_0^1\frac{nx^{n-1}}{1+x}dx=\frac{1}{2}.$$
A: 
I thought it might be instructive to present an approach that is motivated by heuristic argument.  The rigorous proof presented here is more involved than one using by integration by parts, but demonstrates a general approach that can be used to tackle a wide variety of problems in this class.


Note that $\displaystyle \lim_{n\to \infty}(nx^{n-1})=0$ for $\displaystyle 0\le x<1$.  Hence, for any $\displaystyle 0<\delta<1$, we have
$$\lim_{n\to\infty}\int_0^{1-\delta} \frac{nx^{n-1}}{1+x}\,dx=0$$ 
Heuristically, this means that if $\displaystyle \lim_{n\to\infty}\int_0^1 \frac{nx^{n-1}}{1+x}\,dx>  0$, then all of the contribution to its value occurs in the integration near $\displaystyle x=1$.  And near $\displaystyle x=1$, we have $\displaystyle \frac1{1+x}\approx \frac12$.  Therefore, we could guess that 
$$\lim_{n\to \infty}\int_0^1 \frac{nx^{n-1}}{1+x}\,dx=\frac12\lim_{n\to \infty}\int_0^1 nx^{n-1}\,dx=\frac12\tag1$$
We will now prove $(1)$, without using the very straightforward integration by parts approach.

For any given $\epsilon>0$, there exists a number $\delta>0$ such that $\frac{1-x}{2(1+x)}$ whenever $0<1-x<\delta$.  So, for a given $\epsilon>0$, we fix $\delta>0$ accordingly.  Then, we write
$$\begin{align}
\left|\int_0^1\frac{nx^{n-1}}{1+x}\,dx-\frac12\right|&=\left|\frac12\int_0^1\left(\frac{1-x}{1+x}\right)nx^{n-1}\,dx\right|\\\\
&=\frac12 \int_0^{1-\delta}\left(\frac{1-x}{1+x}\right)nx^{n-1}\,dx\\\\
&+\frac12 \int_{1-\delta}^1\left(\frac{1-x}{1+x}\right)nx^{n-1}\,dx\\\\
&\le \frac{\delta}{2(2+\delta)}(1-\delta)^n+\epsilon \left(1-(1-\delta)^n\right)
\end{align}$$
As $n\to \infty$, we see that 
$$\lim_{n\to\infty}\left|\int_0^1\frac{nx^{n-1}}{1+x}\,dx-\frac12\right|<\epsilon$$
Since $\epsilon>0$ is arbitrary, we conclude that 
$$\lim_{n\to\infty}\int_0^1\frac{nx^{n-1}}{1+x}\,dx-\frac12=0$$
And we are done.
A: Using $$\displaystyle 2x<1+x<2\Rightarrow \frac{1}{2x}<\frac{1}{1+x}<\frac{1}{2x}$$
So $$\int^{1}_{0}\frac{nx^{n-1}}{2}dx<\int^{1}_{0}\frac{nx^{n-1}}{1+x}dx<\int^{1}_{0}\frac{nx^{n-1}}{2x}dx$$
So using Sequeeze Theorem $$\lim_{n\rightarrow \infty}\int^{1}_{0}\frac{nx^{n-1}}{1+x}dx=\frac{1}{2}.$$
