How to find $\lim\limits_ {n\to\infty}n^5\int_n^{n+2}\frac{{x}^2}{{ {2+x^7}}}\ dx$? How to find $$\displaystyle\lim_ {n\to\infty}n^5\int_n^{n+2}\dfrac{{x}^2}{{ {2+x^7}}}\ dx$$Can I use Mean Value Theorem? Someone suggested I should use Lagrange but I don't know how it would help.
 A: $$\lim_{n\to\infty} n^5\int_n^{n+2} \frac{x^2}{x^7+2}\operatorname{dx} = \lim_{n\to\infty} \int_n^{n+2}\frac{n^7x^2}{n^2(x^7+2)}\operatorname{dx} $$
$$\lim_{n\to\infty} \int_n^{n+2} \frac{n^5x^{-5}\operatorname{dx}}{1+2x^{-7}}$$
$$\lim_{n\to\infty} \int_n^{n+2} n^5x^{-5}(1-2x^{-7})\operatorname{dx}$$
Can you do it now?
A: $$\frac{x^2}{2+x^7}=\frac1{x^5}-\frac 2{x^5(2+x^7)}$$ and
$$\int_n^{n+2}\frac{x^2}{2+x^7}dx=\frac1{4n^4}-\frac1{4(n+2)^4}-\int_n^{n+2}\frac 2{x^5(2+x^7)}dx.$$
The integral on the right is of order $n^{-11}$ and can certainly be neglected.
Then reducing to the common denominator and expanding
$$\frac{n^5}4\frac{(n^4+8n^3+\cdots)-n^4}{n^4(n+2)^4}\to\color{green}2.$$
A: $$
\begin{align}
2n^5\frac{n^2}{2+(n+2)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le2n^5\frac{(n+2)^2}{2+n^7}\\[12pt]
\frac2{\frac2{n^7}+\left(1+\frac2n\right)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le\frac{2\left(1+\frac2n\right)^2}{\frac2{n^7}+1}
\end{align}
$$
Apply the Squeeze Theorem.
