How to compute this limit using covergence theorem

Question

Can anyone help me compute this limit?

$$\lim_\limits{n \rightarrow \infty}{ \int_\limits{0}^{1}\frac{1+nx^2}{(1+x^2)^n}dx}$$

Answer 1

Note that $(1+x^{2})^{n}\geq 1+nx^{2}+\dfrac{n(n-1)}{2}x^{4}$, hence $\dfrac{1+nx^{2}}{(1+x^{2})^{n}}\leq\dfrac{1+nx^{2}}{1+nx^{2}+(n(n-1)/2)x^{4}}\leq\dfrac{1+nx^{2}}{1+(1/2)(n-1)^{2}x^{4}}$, for $x\in(0,1]$, $\dfrac{1+nx^{2}}{(1+x^{2})^{n}}\rightarrow 0$. Now we have also that $\dfrac{1+nx^{2}}{(1+x^{2})^{n}}\leq\dfrac{1+nx^{2}}{1+nx^{2}}=1$, the limit of integral is $0$, which follows by Lebesgue Dominated Convergence Theorem.

Answer 2

Use the Bernoulli's inequality

$$(1+y)^n\ge 1+ny.$$

Since $x\in [0,1]$ you have:

$$ \int_\limits{0}^{1}\frac{1+nx^2}{(1+x^2)^n}dx\le \int_\limits{0}^{1}\frac{(1+x^2)^n}{(1+x^2)^n}dx=\lim_{n\rightarrow \infty} \int_0^1dx=1 \quad \forall n\in\Bbb N$$

So you can take the limit under the sign of integral, so

$${ \int_\limits{0}^{1}\lim_\limits{n \rightarrow \infty}\frac{1+nx^2}{(1+x^2)^n}dx}=\int_0^10dx=0$$