# Is $\lim_n (1/n)\log(x_n)=\lim_n(1/n)\log(x_n+1)$?

Let $(x_n)$ be a sequence such that $\lim_{n\to\infty}\frac{1}{n}\log(x_n)=L$ with $L<\infty$. My (maybe silly) question is, whether $$\lim_{n\to\infty}\frac{1}{n}\log(x_n+1)=L?$$

Sad, but I do not know how to prove/ disprove that therefore please give me some help.

If $\lim_{n\to\infty}\frac1n \log(x_n) = L$, then $\log(x_n)\sim Ln$ as $n\to\infty$. Therefore $x_n\to\infty$ as $n\to\infty$. Hence $${x_n\over 1 + x_n}\to 1$$ as $n\to\infty$, provided $L > 0$.
• Does this mean that $x_n\sim 1+x_n$, thus $\log(x_n)\sim \log(x_n+1)$, so that my question is to answer with yes in case $L>0$? – math12 Apr 30 '15 at 18:59
• @ncmathsadist I took the liberty of inserting the previously missing term $1/n$ from the first limit expression. I hope that you don't mind. And +1 – Mark Viola Apr 30 '15 at 19:28
Take $x_n=e^{-n}$, then $\lim_{n\to\infty}\frac{1}{n}\log(x_n)=-1$ but $\lim_{n\to\infty}\frac{1}{n}\log(x_n+1)=0$.
• This happens whenever $L < 0$. – ncmathsadist Apr 30 '15 at 18:48
• What, if $L=0$? – math12 Apr 30 '15 at 19:09