1
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

Hint: $\ln(n-1)-\ln(n)=\ln\frac{n-1}{n}$. What is $\lim_{n\to\infty} \frac{n-1}{n}$?

$\endgroup$
5
$\begingroup$

$$0\lt\ln(n)-\ln(n-1)=\int_{n-1}^n{dx\over x}\lt\int_{n-1}^n{dx\over n-1}={1\over n-1}$$

Now squeeze.

$\endgroup$
1
$\begingroup$

$$\lim_{x\rightarrow \infty }\ln(1-1/n)=\ln(1)=0$$

$\endgroup$
1
$\begingroup$

Hint: $$\log a -\log b= \log \frac{a}{b}$$ Which leads to $$\lim_{n\to\infty} \log \frac{n+1}{n}=\lim \log\left(\frac{1+\frac 1n}{1}\right)=\log\left(\lim 1+\frac 1n\right)=\log 1=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.