<s>[INST] A Question Lebesgue Dominated Convergence Theorem 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.
 [/INST] 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.
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