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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.

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  • $\begingroup$ @Zacky Could you please show me how to do this and give me some explanations? I have seen this technique used, but I can't really grasp it. P. S. : I know what L'H is, but I don't know how to find the limit of the integral itself. $\endgroup$ – JoMath Jun 10 at 13:13
  • $\begingroup$ Thank you! Just one more question to make sure that I fully understand this : $\lim\limits_{n\to \infty} (F(2n)-F(n)) =0$ because it is basically $F(\infty) - F(\infty)$? $\endgroup$ – JoMath Jun 10 at 13:23
  • $\begingroup$ Then the numerator's limits is $0$ because both $n$ and $2n$ $\to \infty$? $\endgroup$ – JoMath Jun 10 at 13:42
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    $\begingroup$ @Zachy, Unless you actually show that $\lim \int_n^{2n} \frac{x}{1+x^5} dx = 0$ your solution is not complete. Depending on the integrand function, it is not necessarily true that $\int_n^{2n} (\cdots) \to 0$. You should also edit to correct your derivative of $1/n^3$. $\endgroup$ – PierreCarre Jun 10 at 19:05
  • $\begingroup$ Sorry @Zacky I misspelled your username in the previous comment. $\endgroup$ – PierreCarre Jun 10 at 20:17
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$$\lim\limits_{n\to \infty} \left(n^3 \int_{n} ^{2n}\frac{x dx} {1+x^5}\right) $$

Put $x=nt$ and $dx=ndt$

$$\lim_{n\rightarrow \infty}n^5\int^{2}_{1}\frac{t}{1+n^5t^5}dt$$

$$\int^{2}_{1}\lim_{n\rightarrow \infty}\frac{n^5t}{1+t^5n^5}dt=\int^{2}_{1}t^{-5}dt$$

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    $\begingroup$ Small typo last line. it should be n^5*t and the last integral should be t^-4. Otherwise, Very nice solution $\endgroup$ – user600016 Jun 10 at 17:12
  • $\begingroup$ @jacky You could add a comment on the correctness/possibility of exchanging limit and integral. $\endgroup$ – PierreCarre Jun 11 at 6:53
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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 $g'(n)$.

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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 $n \rightarrow \infty$.

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  • $\begingroup$ Simple approach. +1 $\endgroup$ – Paramanand Singh Jun 10 at 15:44
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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'Hôpital'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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