I have a question about the application of Lebesgue Dominated Convergence Theorem.
$\lim _{n \rightarrow \infty} \int_{0}^{1}\left(1+n x^{2}\right)\left(1+x^{2}\right)^{-n} d x=?$
Firstly i have a reference about this question http://www.ma.man.ac.uk/~mdc/old/341/solutions3.pdf
By using this reference my solution is
Starting from $(1+n x^2)(1+x^2)>1+(n+1) x^2$ to see that $1+n x^2>\frac{1+(n+1) x^2}{1+x^2}$ and so; $$ \frac{1+n x^2}{(1+x^2)^{n}}>\frac{1+(n+1) x^2}{(1+x^2)^{n+1}} $$
For $x=0$ then all terms in the sequence of (fn(x)) equal 1 so the limit is $1 .$ If $0<x<1$ and $x=1$ we start with the observation that the binomial expansion gives $(1+x^2)^{n} \geq 1+n x^2+\frac{n(n-1)}{2} x^{4}$ and so $$ \frac{1+n x^2}{(1+x^2)^{n}} \leq \frac{1+n x^2}{1+n x^2+\frac{n(n-1)}{2} x^{4}} \rightarrow 0 $$ as $n \rightarrow \infty$. Thus the limit is 1 if $x=0$ and 0 elsewhere, that is, 0 a.e. $(\mu)$
We could choose the dominating function to be the n = 3 term, $h(x)=(1+3 x^2) /(1+x^2)^{3}$ so $h$ is integrable. But also since the sequence of functions is decreasing each function, at least for $n \geq 3,$ is integrable. Hence, using the Dominated convergence Theorem we can justify the interchange in $$ \lim _{n \rightarrow \infty} \int_{0}^{1} \frac{1+n x^2}{(1+x^2)^{n}} d x=\int_{0}^{1} \lim _{n \rightarrow \infty} \frac{1+n x^2}{(1+x^2)^{n}} d x=\int_{0}^{1} 0 dx=0 $$
I want to ask about my solution . is it right or not? Or is there any missing solution? Thanks for your support.
You can make this easier: since $(1+x^2)^n \ge 1 +nx^2,$ we have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2} =1.$$
So the constant function $1,$ which is certainly integrable on $[0,1],$ is a dominating function for your sequence.
As you said, we also have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2+[n(n-1)]/2\cdot x^4} \to 0$$
on the interval $(0,1].$ By the DCT we're done.