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Having a hard time finding this limit, it must be some natural log replacement I figured but I quite don't know how to replace it as such

$\lim _{n\to \infty }\left(n\int _0^1\:\left(\frac{x^n}{\left(x^n+x+1\right)}dx\right)\right)$

Managed to see that it can be written as almost the natural log by doing this thing here:

$\lim _{n\to \infty }\left(\int _0^1\left(\frac{\left(nx^{n-1}+1-1\right)}{\left(x^n+x+1\right)}dx\right)\:\:\right)$

$\lim _{n\to \infty }\left(\left(ln\left(3\right)-\int _0^1\left(\frac{1}{\left(x^n+x+1\right)}dx\right)\right)\:\:\right)$

Still have to solve this one: $\int _0^1\left(\frac{1}{\left(x^n+x+1\right)}dx\right)$

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  • $\begingroup$ I think so there is mistake in 3rd line. $\endgroup$
    – RAHUL
    Oct 31, 2021 at 19:14
  • $\begingroup$ @RAHUL it was, ln(3) $\endgroup$
    – Alex Mihoc
    Oct 31, 2021 at 19:19
  • $\begingroup$ Yaa, so that integrand is only left to solve $\endgroup$
    – RAHUL
    Oct 31, 2021 at 19:22
  • $\begingroup$ You can remove the $n$ at the front and bring it in. Note that as $n\to\infty$, $nx^{n}=nx^{n-1}$ $\endgroup$
    – RAHUL
    Oct 31, 2021 at 19:26
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    $\begingroup$ $$\int_{0}^{1}\frac{dx}{x^{n}+x+1}$$ converges to $0.69..$ as $n$ approach $\infty$ $\endgroup$
    – RAHUL
    Oct 31, 2021 at 19:34

1 Answer 1

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Here is the solution I found it eventually! I had to do a substitution...enter image description here

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  • $\begingroup$ Use mathjax tools for formatting. By the way, glad you got it. $\endgroup$
    – RAHUL
    Nov 1, 2021 at 8:36

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