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Let $$ f(x)=1+\frac{\log x}{x}+o\left(\frac{\log x}{x}\right), x\to\infty. $$

Do we then have $$\lim_{x\to\infty}x\log(f(x))=\infty?$$

My answer would be yes, since:

First of all, I think that we have $$ f(x)\sim 1+\frac{\log x}{x}, x\to\infty~~~(1) $$ because $f(x)- (1+\frac{\log x}{x})=o\left(\frac{\log x}{x}\right)$ and for $g(x)=o\left(\frac{\log x}{x}\right)$ we have $g(x)\in o\left(1+\frac{\log x}{x}\right)$.

Thus, we have $$ \log(f(x))\sim\log\left(1+\frac{\log x}{x}\right)=\frac{\log x}{x}+O\left(\left(\frac{\log x}{x}\right)^2\right), x\to\infty $$

Consequently,

$$ x\log(f(x))\sim \log x+O\left(\frac{\log^2(x)}{x}\right), x\to\infty $$

and hence $$ \lim_{x\to\infty}x\log(f(x))=\lim_{x\to\infty}\left(\log x+O\left(\frac{\log^2(x)}{x}\right)\right)=\infty, $$ since each function $h(x)\in O\left(\frac{\log^2(x)}{x}\right)$ tends to $0$ as $x\to\infty$.

Would you agree?

Astyx
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Salamo
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  • What is your definition of $\sim$ ? – Astyx Feb 26 '18 at 14:49
  • I think I should use that $f(x)\sim g(x)$ if and only if $f(x)-g(x)=o(g(x)), x\to\infty$ what here should be equivalent to $\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$. – Salamo Feb 26 '18 at 14:51
  • Not only is $f(x) \sim 1 + \frac{\log x}{x}$, but also $f(x) \sim 1 + 1/x$ and $f(x) \sim 1 + e^{-x}$ and even $f(x) \sim 1$. The notion of $\sim$ is definitely not the one you want to start with here. – Antonio Vargas Feb 26 '18 at 19:21

1 Answers1

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We have that $${\log x \over x} \longrightarrow_{x\to \infty} 0$$ and $\log(1+y)\sim y$ as $y\to 0$. This means that $$\log(f(x)) = \log\left(1+{\log x\over x} + o\left({\log x\over x}\right)\right) \sim {\log x\over x}+o\left({\log x\over x}\right) \sim {\log x\over x}$$

Consequently $$x\log(f(x))\sim \log x \to \infty$$

Note that $1 + {\log x\over x} \sim 1$, so writing $f(x) \sim 1+ {\log x\over x} $ is not strong enough to get the result you want.

Astyx
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  • But we also have that $f(x)= 1+\frac{\log x}{x}+o\left(\frac{\log x}{x}\right)\sim 1$? – Salamo Feb 26 '18 at 15:16
  • Not the last $\sim$, at least not as such. I simply use that ${\log(x)\over x} = o(1)$ and that $\log(1+y)\sim y$ as $y\to 0$ – Astyx Feb 26 '18 at 15:19
  • But $f(x)\sim 1+\frac{\log x}{x}, x\to\infty$ is not wrong, isn't it? – Salamo Feb 26 '18 at 15:44
  • I just do not understand your sentence that $f(x)\sim 1+\frac{\log x}{x}$ is not strong enough because it implies that $\log(f(x))\sim\log\left(1+\frac{\log x}{x}\right)\sim \frac{\log x}{x}$ and hence $x\log(f(x))\sim \log x$ and thats what I want to get. – Salamo Feb 26 '18 at 15:47
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    It doesn't imply what you say. You cannot take the log of equivalents. It is not generally true that $f(x)\sim 1+{\log x\over x}\implies \log(f(x)) \sim \log(1+{\log x \over x})$. For instance $1+{1\over x}\sim 1+{\log x\over x}$ – Astyx Feb 26 '18 at 16:09
  • Ok, but the following link claims that if $f\sim g$, then $\log f\sim \log g$. https://en.m.wikipedia.org/wiki/Asymptotic_analysis – Salamo Feb 26 '18 at 16:16
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    That's wrong for f close to 1 at infinity : take 1+1/x and 1+1/x^2 as a counterexample – Astyx Feb 26 '18 at 16:22
  • Maybe I repeat myself, but for our specific $f(x)$, I do not see the problem, why $\log(f(x))=\log(1+\frac{\log x}{x}+g(x))\sim\log(1+\frac{\log x}{x})$ for $g(x)=o(\log x/x)$ should not be satisfied, – Salamo Feb 26 '18 at 18:01
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    @Salamo It is satisfied, it's your justification that is flawed. – Antonio Vargas Feb 26 '18 at 19:23