# Natural log limit question

I have to find $$\lim_{n\to\infty}\left(\ln(n-1)-\ln(n)\right)$$

I'm pretty sure I need to solve this using the asymptotes. So if I use the rule for logs I can do lim (ln((n-1)/n)) and I know that n can't equal 0 because then the fraction would be undefined. I'm just confused on how to evaluate the limit. How do I find the horizontal asymptote?

Hint: $\ln(n-1)-\ln(n)=\ln\frac{n-1}{n}$. What is $\lim_{n\to\infty} \frac{n-1}{n}$?
$$0\lt\ln(n)-\ln(n-1)=\int_{n-1}^n{dx\over x}\lt\int_{n-1}^n{dx\over n-1}={1\over n-1}$$
$$\lim_{x\rightarrow \infty }\ln(1-1/n)=\ln(1)=0$$
Hint: $$\log a -\log b= \log \frac{a}{b}$$ Which leads to $$\lim_{n\to\infty} \log \frac{n+1}{n}=\lim \log\left(\frac{1+\frac 1n}{1}\right)=\log\left(\lim 1+\frac 1n\right)=\log 1=0$$