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From the taylor's theorem, $\ln(1+x) = \frac{1}{1+\xi}\cdot x$, where $\xi$ is between $0$ and $x$.

Then $\ln(1+\frac{1}{x})$ = $\frac{1}{1+\xi}\cdot\frac{1}{x}$.

So, $\ln(1+\frac{1}{x})$ = $O$($\frac{1}{x}$).

Is this right? And in line 2, does $\xi$ remains unchanged?

Not related: Also i stumbled onto this solution online, http://imgur.com/a/POAYD. i find the use of limit strange.

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    For $x -> 0$, $\ln(1+\frac{1}{x})$ = $O(\frac{1}{x})$. Is that right? – 2017-02-13
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    Oh, sorry, yes, but this time $x$ tends to $\infty$ – 2017-02-13
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    Then does $\xi$ remains unchanged when i substitute $\frac{1}{x}$ for $x$? – 2017-02-13
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    I am confused a bit, I do not think it remains. The $\xi$ is usually used for a series around a finite point. – 2017-02-13
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    what should $\xi$ be then after the substitution? – 2017-02-13
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    The big-O and little-o notation don't mean anything unless you specify where $x$ is going. Otherwise it is similar to writing $\lim f(x)$ without saying where $x$ is going. E.g $\tan x=x+O(x^3)$ as $x\to 0.$ But as $x\to \pi /2$ we have $\tan x \ne x+O(x^3).$ – 2017-02-13
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    Is your desire to prove the Taylor Series in general, to prove the limit in the post you link, or to know if $\frac{1}{1+\xi} \frac 1x = O(\frac 1x)$ – 2017-02-13
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    @BrevanEllefsen The 3rd option. Also im wondering what will be the new $\xi$ be. I have seen some materials where they makes no changes to $\xi$ after substituting the independent variable of the function as another form. – 2017-02-14
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    @user254665 I used the Taylor's theorem to write Taylor series in the finite form. The last term is the product of a constant and a specific term of the Taylor series, hence the last term is $O$(term). – 2017-02-14
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    @Peter i hope someone can give me the correct explanation soon :( – 2017-02-14

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No. You can't do this because $\xi$ is a function of $x$. Moreover, when using Big-O notation we must be talking about asymptotic around a point (possibly $\infty$) and that isn't the case here. We say that $f(x) \in O(g(x)) $ (also written as $f(x) =O(g(x)) $) when $\displaystyle\limsup_{x\to a} \left|\frac{f(x)}{g(x)}\right| < \infty$.

On the other hand, around the origin we have that $\log(1+x) = x + O(x^2)$, which can clearly be written as $\log(1+x) = O(x)$. Let $x \to \frac{1}{x}$ and we get your result, at least for all $|x| < 1$

Also, note that the two instances of $\xi$ are not the same. In the first case we have $$\log(1+x) = \frac{x}{1+\xi} \implies \xi = \frac{x}{\log(1+x)}-1$$ And in the second case we have $$\log(1+1/x) = \frac{1/x}{1+\xi}\implies \xi=\frac{1}{x\log(1+1/x)}-1$$

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    The $O$ notation is widely used. I think u are referring to the rate of convergence, which is not related to my question. – 2017-02-14
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    @LittleRookie I'm quite positive I know what Big-O notation is. It refers to asymptotic growth of a function around a given point, and is defined in terms of a limit superior – 2017-02-14
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    So what should the proper form of $\xi$ be? – 2017-02-14
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    @LittleRookie I don't get what you mean. $\xi$ depends on the $x$ we choose — it's not a constant for all $x$ and so we can't use the constant rule for Big O notation. It's a function, not a constant. – 2017-02-14
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    Indeed, $\xi$ depends on $x$, to be exact with my question, what does $\xi$ depends on now? – 2017-02-14
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    @LittleRookie No, they are not the same $\xi$. I have laid this out. – 2017-02-14
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    the different Landau notation in it general form (included the big Oh) cant be stated using a limit. However in many cases is true that it can be stated as a limit. Take a look [here](https://en.wikipedia.org/wiki/Big_O_notation#Formal_definition). – 2017-06-07
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    @Masacroso while the Wikipedia page isn't extremely clear on the conditions where the two are not equal, I do agree with you - they may not be equivalent in general. Nevertheless, I don't see any reason why they would differ in this case – 2017-06-07