Is the limit of $f(x)-\ln(x)$ equal to the limit of $f(x+n)-\ln(x)$? Given that $\displaystyle \lim_{x \to +\infty} (f(x) - \ln x) = 0$, can I say that $\displaystyle \lim_{x \to +\infty} (f(x+n)- \ln x) = 0$?
 A: 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.
A: Yes, the limit is 0.
$$\lim_{n \to \infty} (f(x+n)- \ln(n))$$
$$= \lim_{n \to \infty} (((f(x+n)- \ln(x+n)) + (\ln(x + n) - \ln(n)))$$
$$= \lim_{n \to \infty}  (\ln(x + n) - \ln(n))$$
$$= \lim_{n \to \infty} (\ln \frac{x + n}{n})$$
$$= \lim_{n \to \infty} (\ln{1})$$
$$= 0$$
A: $f(x+n)-ln(x)=f(x+n)-ln(x+n-n)=f(x+n)-ln((x+n)(1-{n\over{x+n}}))$
$=f(x+n)-ln(x+n)-ln(1-{n\over{x+n}})$.
$lim_{x\rightarrow +\infty}f(x)-ln(x)=0$ 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.
