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Given that $\displaystyle \lim_{x \to +\infty} (f(x) - \ln x) = 0$, can I say that $\displaystyle \lim_{x \to +\infty} (f(x+n)- \ln x) = 0$?

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    Yes because $\log(x + n) - \log x \to 0$ as $x \to \infty$ – 2017-01-04
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    Well don't mind the letters, I think it is the same limit whether I change $x$ for $n$. – 2017-01-04
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    The way you have written the question $x$ and $n$ play different roles. $x$ is the variable which is used for limit operation and $n$ is a fixed constant. – 2017-01-04
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    $n$ is a constant....so,increment of $n$ of $f(x)$ as $x\rightarrow\infty$ is negligible... – 2017-01-04
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    Sorry I wasn't answering your comment. And I meant that $\lim_n (f(n+x)- \ln n)$ should equal $\lim_x (f(x+n)- \ln x)$. – 2017-01-04
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    Sure tatan, but I don't know how to prove it properly. How can I introduce the $\log (x+n)$ in my equation? – 2017-01-04
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    Please edit your question to reflect what you say in comments and then it would be easy to give a good answer. – 2017-01-04

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Yes, if $n$ is constant.

\begin{align}f(x+n)-\ln {x}&=f(x+n)\pm\ln{(x+n)}-\ln x\\[0.2cm]&=f(x+n)-\ln {(x+n)}+(\ln{(x+n)}-\ln {(x)})\end{align} So, this reduces to whether $\lim_{x\to\infty}\ln{(x+n)}-\ln{(x)}=0$ or not. Due to continuity of $\ln$ $$\lim_{x\to\infty}\ln{(x+n)}-\ln{(x)}=\lim_{x\to \infty}\ln {\left(\frac{x+n}{x}\right)}=\lim_{x\to \infty}\ln {\left(1+\frac{n}{x}\right)}=\ln {(1+0)}=0$$ If you mean $x=n$ and $n$ varies as $x$, in other words if $n$ is not a constant, then the last equation becomes $$\lim_{x\to\infty}\ln{(x+x)}-\ln{(x)}=\lim_{x\to \infty}\ln {\left(\frac{x+x}{x}\right)}=\ln 2$$ and the answer in this case is no.

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    your very first line seems to be wrong...how should $$f(x+n)-\ln x=f(x+n)\pm\ln(x+n)-\ln x$$ hold? – 2017-01-04
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    @user190080 $\pm \ln{(x+n)}=\ln{(x+n)}-\ln{(x+n)}=0$. – 2017-01-04
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    @JimmyR. ah...well...ok then :) – 2017-01-04
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Yes, the limit is 0.

$$\lim_{n \to \infty} (f(x+n)- \ln(n))$$

$$= \lim_{n \to \infty} (((f(x+n)- \ln(x+n)) + (\ln(x + n) - \ln(n)))$$

$$= \lim_{n \to \infty} (\ln(x + n) - \ln(n))$$

$$= \lim_{n \to \infty} (\ln \frac{x + n}{n})$$

$$= \lim_{n \to \infty} (\ln{1})$$

$$= 0$$

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    My answer was for the original question, before the edit. – 2017-01-04
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    Sorry for the edit. Well I just changed $x$ for $n$ and $n$ for $x$ so it remains the same. – 2017-01-04
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$f(x+n)-ln(x)=f(x+n)-ln(x+n-n)=f(x+n)-ln((x+n)(1-{n\over{x+n}}))$

$=f(x+n)-ln(x+n)-ln(1-{n\over{x+n}})$.

$lim_{x\rightarrow +\infty}f(x)-ln(x)=0$ implies that $lim_{x\rightarrow +\infty}f(x+n)-ln(x+n)=0$. But $lim_{x\rightarrow +\infty}ln(1-{n\over{x+n}})=0$. So the answer is yes.