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The question is somewhat related to the discussions I had about this other post. It seems for every integer $n>1$ the limit $$\lim_{x\to\infty}\left(\frac{x+n}{\ln(x+n)}-\frac x {\ln x}\right)$$ does exist. But I wasn't able to obtain a closed form for it (I admit that I didn't try enough). So, is there a closed form for it?

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  • $\begingroup$ What makes you think that the limit is not zero? $\endgroup$ – b00n heT Oct 18 '16 at 11:15
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Hint. By the Mean Value Theorem, $$\frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}=((x+n)-x)\cdot\frac{d}{dx}\left(\frac{x}{\ln x}\right)_{x=t}=n\left(\frac{1}{\ln t}-\frac{1}{\ln^2 t}\right)$$ with $x<t<x+n$. Now $t$ goes to infinity as $x\to\infty$.

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  • $\begingroup$ The last part isn't really satisfactory, since we don't really know $t$... It's easier to note that $|n\left(\frac{1}{\ln t}-\frac{1}{\ln^2 t}\right) | \le n \left( \frac{1}{\ln x} + \frac{1}{(\ln x)^2} \right)$, and now this obviously converges to zero. $\endgroup$ – Najib Idrissi Oct 18 '16 at 11:31
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    $\begingroup$ @Najib Idrissi. We do not know $t$, but we know that $t>x$. Hence $x\to +\infty$ implies that $t\to +\infty$. $\endgroup$ – Robert Z Oct 18 '16 at 11:34
  • $\begingroup$ I'm not questioning the fact that your proof is correct, but a beginner could wonder whether you're really allowed to do what you did -- I just gave an alternative way of seeing the result. $\endgroup$ – Najib Idrissi Oct 18 '16 at 11:36
  • $\begingroup$ Another question just popped into my mind, which I am too lazy to ask separately. In my question, what if we replace $n$ by $\ln x$? Maybe I'm wrong but, the mean value theorem doesn't look much promising in this case. It would be nice if you cover this case as well. Although I have consumed the only upvote that I could give, and don't have anything more to offer. $\endgroup$ – polfosol Oct 18 '16 at 12:10
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    $\begingroup$ In that case the limit is 1 because $\frac{x}{\ln(x+\ln(x))}-\frac{x}{\ln x}\to 0$ and $\frac{\ln(x)}{\ln(x+\ln(x))}\to 1$. $\endgroup$ – Robert Z Oct 18 '16 at 12:18
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see my answer $$\lim_{x\to\infty} \frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}$$ $$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x\left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$ $$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x+\ln \left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$ $$=\frac{x\left(1+0\right)}{\ln x+\ln \left(1+0\right)}-\frac{x}{\ln x}$$ $$=\frac{x}{\ln x}-\frac{x}{\ln x}=0$$ the limit exists for all real n

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  • $\begingroup$ I vaguely remember that we were not allowed to treat $\infty-\infty$ like this. The final answer seems right nonetheless $\endgroup$ – polfosol Oct 18 '16 at 11:21
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    $\begingroup$ This is nonsense. A similar "proof" could "show" that the limit of $x(\ln(x+1) - \ln(x))$ would be zero as $x \to \infty$, even though the limit is actually $1$... You can't just take the limit of part of the expression, you have to take the limit of everything at once. $\endgroup$ – Najib Idrissi Oct 18 '16 at 11:27

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