I want to compute that $\lim\limits_{n\to\infty}\int_{[0,1]}\frac{nx^{n-1}}{1+x} \, d\lambda(x)=\frac{1}{2}$, where $\lambda$ is the Lebesgue measure.
I want to use Lebesgue dominated convergence theorem, but I can't find a uniform bound for $|\frac{nx^{n-1}}{1+x}|$, and the limit function tends to be $\begin{cases}0 & x<1 \\\infty & x=1\end{cases}.$ How do I deal with this integral?
On
Make the substitution of variable $u = x^n$ which implies $du = n x^{n-1}dx$ to get $$\int_0^1 \frac{nx^{n-1}}{1+x} \ dx = \int_0^1 \frac{1}{1+ u^{1/n}} \ du$$
And now, you can apply Lebesgue dominated convergence theorem as $$0 \le \frac{1}{1+ u^{1/n}} \le 1$$ and $\lim\limits_{n \to \infty} \frac{1}{1+ u^{1/n}} = 1/2$ for $u \in (0,1]$.
Dominated convergence allows to interchange limit and integral.The integral of the pointwise limit is zero. However, the limit of the integrals is non-zero. So dominated convergence cannot work for this example.