Find the value of $\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{x+1}dx$ Find the value of $$\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{x+1}dx$$
I just found that limit converge in $0<L<1$ but I don't know what should I do next.
I tried $\frac{nx^{n-1}}{x+1}<nx^{n-1}$ to sqeeze but It wasn't helpful.
 A: Write the integral as $x^{n}\frac 1 {1+x}|_0^{1}+\int_0^{1} \frac {x^{n}} {(1+x)^{2}}dx$ and note that the integral in the second term is $\leq \int_0^{1}x^{n}dx=\frac 1 {n+1} \to 0$. Hence the limit is $\frac 1 2$.
A: Use $1\le x+1\le 2$. So the part with $1/(x+1)$ is easily controlled. Then integrate $nx^{n-1}$ between $0,1$. So we have the bounds $1/2$ and $1$ for the limit, now we have to refine. Partial integration:
$$
\begin{aligned}
J_n&:=\int_0^1 \frac{nx^{n-1}}{x+1}\; dx
\\
&=
\int_0^1 (x^n)'\cdot\frac1{x+1}\; dx
\\
&=
\left[\ (x^n)'\cdot\frac1{x+1}\ \right] _0^1
-
\int_0^1 x^n\cdot\frac{-1}{(x+1)^2}\; dx
\ .
\end{aligned}
$$
The first term is $1/2$. The second one can be now estimated as above, goes to zero.
Computer check:
? J(n) = intnum(x=0, 1, n*x^(n-1)/(x+1) );
? \p100
? J(10^10)
%13 = 0.5000000000272192536894630414405030284996885838506060285022828092024169433977769922283503423833021872

A: $$I=\lim_{n\rightarrow \infty}\int^{1}_{0}\frac{nx^{n-1}}{1+x}dx$$
Put $x^n=t$ and $nx^{n-1}dx=dt$
$$\lim_{n\rightarrow \infty}\int^{1}_{0}\frac{1}{t^{\frac{1}{n}}+1}dt$$
Using Dominance Convergence Theorem
$$\int^{1}_{0}\lim_{n\rightarrow \infty}\frac{1}{1+t^{\frac{1}{n}}}dt=\int^{1}_{0}\frac{1}{2}dt=\frac{1}{2}.$$
A: Integrate by parts
$$\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{x+1}dx = \lim_{n\to\infty}  \frac{nx^{n}/n}{x+1}|_0^1 + \int_0^1 \frac{nx^{n}/n}{(x+1)^2}dx  = 1/2+\lim_{n\to\infty}  \int_0^1 \frac{x^{n}}{(x+1)^2}dx = 1/2$$
The last term is because $$0\le \int_0^1 \frac{x^{n}}{(x+1)^2}dx \le \int_0^1 x^{n}dx = \frac{1}{n+1} \to 0$$
A: Split the integral into two parts.  $I_n=\int_0^{1-u}f_n(x)dx+\int_{1-u}^1f_n(x)dx=I_{n1}+I_{n2}$, where $f_n(x)=\frac{nx^{n-1}}{1+x}$.
$I_{n1}\lt \int_0^{1-u}nx^{n-1}dx=(1-u)^n \to 0$ as $n\to \infty$
$\int_{1-u}^1\frac{nx^{n-1}}{2}dx\lt I_{n2}\lt \int_{1-u}^1 \frac{nx^{n-1}}{2-u}dx$
As $n\to \infty$ we get $\frac{1}{2}\lt lim I_{2n} \lt \frac {1}{2-u}$  We used $\int_{1-u}^1nx^{n-1}dx=1-(1-u)^n\to 1$
Since $u$ can be arbitrarily small, $lim_{n\to \infty}I_n=\frac{1}{2}$
