This is a qualifying exam practice question - so not being graded for homework purposes, just studying!
Calculate $\lim_{n \rightarrow \infty} \int_0^\infty \frac{x^n}{ x^{(n+3)}+1} dx$
I tried the following:
$\lim_{n \rightarrow \infty} \int_0^\infty \frac{x^n}{ x^{(n+3)}+1} \, dx$ = $\frac{d}{dn}\int_0^\infty \int_0^\infty\frac{x^n}{ x^{(n+3)}+1}dn \, dx$ = -$\frac{d}{dn} \int_0^\infty \frac{\ln(x^3+1}{x^3 \ln(x)} \, dx$
Not really sure where to go from here, any advice would be appreciated!
On
Hint. Note that $$\int_0^{\infty}\frac{x^n}{ x^{n+3}+1}\,dx=\int_0^{1}\frac{x^n}{ x^{n+3}+1}\,dx+\int_1^{\infty}\frac{dx}{x^3}-\int_1^{\infty}\frac{dx}{x^3( x^{n+3}+1)},$$ where $$0\leq \int_0^1\frac{x^n}{ x^{n+3}+1}\,dx \leq \int_0^1 x^n \,dx=\left[\frac{x^{n+1}}{n+1}\right]_0^{1}=\frac{1}{n+1},$$ and $$0\leq \int_1^{\infty}\frac{dx}{x^3( x^{n+3}+1)}\leq \int_1^{\infty}\frac{1}{ x^{n+3}}\,dx=\left[-\frac{1}{(n+2)x^{n+2}}\right]_1^{\infty}=\frac{1}{n+2}.$$
Hint:
You have $$\lim_n \int_0^\infty \frac{x^n}{x^{n+3}+1} = \lim_n \int_0^1 \frac{x^n}{x^{n+3}+1}+\lim_n \int_1^\infty \frac{x^n}{x^{n+3}+1}$$
Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^n\rightarrow 0$ so the first term is easy to calculate.
The second converges to $\frac{1}{x^3}$ and then you need to evaluate $\int_1^\infty \frac{1}{x^3} dx$.