Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $ Compute $\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $. I tried to apply the first mean value theorem for  definite integrals and then apply the squeeze theorem, but it didn't work. The answer given by the book is $\frac{7}{24}$,but I can't see how to get to it. 
 A: Hint:
Rewrite this expression as
$$ \frac{\displaystyle\int_{n} ^{2n}\frac{x\,\mathrm dx} {1+x^5}}{\dfrac 1{n^3}}$$
and apply L'Hospital's rule.
For this you have to set 
$$f(y)=\int_{0} ^{y}\frac{x\,\mathrm dx} {1+x^5}$$
and compute the  derivative of the function $g(y)=f(2y)-f(y)$ with the first fundamental theorem of integral calculus.
Last, you'll need  to have an equivalent of $g'(n)$.
A: For $x \in [n,2n]$:
$\dfrac{x}{x^5(1+(1/n)^5)}\le \dfrac{x}{1+x^5} \le \dfrac{x}{x^5(1+(1/2n)^5)}$
Integrate the lower bound:
$L(n):=\dfrac{1}{1+(1/n)^5}\displaystyle{ \int_{n}^{2n}}x^{-4}=$
$\dfrac{1}{1+(1/n)^5}(x^{-3}/(-3))\big ]^{2n}_{n}=$
$\dfrac{1}{1+(1/n)^5}(1/3)(1/n^3-1/(2n)^3)$=
$\dfrac{1}{1+(1/n)^5}\dfrac{7}{24n^3};$
Integrate the upper bound:
$U(n):=\dfrac{1}{1+(1/2n)^5}\dfrac{7}{24n^3}$.
Now consider 
$n^3 L(n)\le n^3\displaystyle{\int_{n}^{2n}}\dfrac{x}{1+x^5}dx \le n^3U(n).$
Take the limit $n \rightarrow \infty$.
A: We should start by noting that 
$$
0 \leq \int_n^{2n}\frac{x}{1+x^5}dx\leq \int_n^{2n} \frac{x}{1+x^4} dx=\left[\frac 12 \arctan(x^2)\right]_n^{2n}=\frac 12 (\arctan(2n)-\arctan n) \to 0.
$$
This justifies that we can apply L'Hospital's rule getting
$$
\lim n^3 \int_n^{2n}\frac{x}{1+x^5}\,dx = \lim \frac{\int_n^{2n}\frac{x}{1+x^5}\,dx}{\frac{1}{n^3}} = \lim \frac{2 \cdot \frac{2n}{1+(2n)^5}-\frac{n}{1+n^5}}{\frac{-3}{n^4}} = \frac{4}{-3 \cdot 32}- \frac{1}{-3 \cdot 1}=\frac{7}{24}.
$$ 
A: $$\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $$
Put  $x=nt$ and $dx=ndt$
$$\lim_{n\rightarrow \infty}n^5\int^{2}_{1}\frac{t}{1+n^5t^5}dt$$
$$\int^{2}_{1}\lim_{n\rightarrow \infty}\frac{n^5t}{1+t^5n^5}dt=\int^{2}_{1}t^{-5}dt$$
