<s>[INST] 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. 
 [/INST] 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
$$
\lim n^3 \int_n^{2n}\frac{x}{1+x^5}\,dx = \lim \frac{\int_n^{2n}\frac{x}{1+x^5}\,dx}{\frac{1}{n^3}} = \lim \frac{2 \cdot \frac{2n}{1+(2n)^5}-\frac{n}{1+n^5}}{\frac{-3}{n^4}} = \frac{4}{-3 \cdot 32}- \frac{1}{-3 \cdot 1}=\frac{7}{24}.
$$ 
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