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Is $$\lim_{n\rightarrow +\infty}\ln\left(\frac{n+1}{n}\right)=0?$$ Because it is $\ln(1+\frac{1}{n})$ and $\frac{1}{n}$ tends to $0$, since $n$ tends to infinity, so the limit becomes $\ln(1+0)=\ln(1)=0$.

Is this right, or is there any remarkable limit related to this?

P.S. I am not used to formatting yet, I didn't really understand the rules, but I did what I could.

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That's right.${}$ –  David Mitra Feb 8 '13 at 21:25
    
There's a related limit $\lim \ln \left(\frac{n+1}{n}\right)^n = 1$. –  sdcvvc Feb 9 '13 at 1:28
    
@sdcvvc well, yes, but that is actually a remarkable limit: $\lim \ln \left(\frac{n+1}{n}\right)^n = 1$ is $ln\ lim\left(1+\frac{1}{n}\right)^n= ln e^{\lim\frac{1}{n}\*n}= ln e^1=1$ –  Bujanca Mihai Feb 10 '13 at 12:29

2 Answers 2

up vote 1 down vote accepted

$\ln$ is continuous at $1$, $\ln 1 = 0$, and $\lim_n \frac{n+1}{n} = 1$, hence $\lim_n \ln(\frac{n+1}{n}) = 0$.

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Yes, continuity of the logarithm lets you bring the limit inside the logarithm so

$$\lim_{n\rightarrow \infty} \ln\left(\frac{n+1}{n}\right)=\ln\left(\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)\right)$$

and yes you can do this for any continuous function besides logarithm.

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