## anonymous 5 years ago Please help me ... limit to infinity of ln (n+1) / ln n

1. anonymous

use l'hopital's rule here ^_^: $\lim_{x \rightarrow \infty} {\frac{1}{n+1}}(n) = \lim_{x \rightarrow \infty} \frac{n}{n} = 1$ ^_^

2. anonymous

did you understand it? :)

3. anonymous

so, Ln (n + 1) / Ln n = Ln [(n+1) / n] , right or wrong? I don't have any clue.

4. anonymous

Since you have something in the form of $$\infty/\infty$$ you need to use l'Hopital's rule. Which is that the limit of the original ratio is the limit of the ratio of the derivatives. if $$(\lim_{n \rightarrow \infty} f(n)= \lim_{n \rightarrow \infty} g(n)) \in \{\infty,0\}$$ $\lim_{n \rightarrow \infty}\frac{f(n)}{g(n)} = \lim_{n \rightarrow \infty} \frac{f'(n)}{g'(n)}$ The derivative of ln(n) = $$\frac{1}{n}$$ The derivative of ln(n+1) = $$\frac{1}{n}$$ So the ratio of the derivatives will be $\frac{\frac{1}{n}}{\frac{1}{n}} = \frac{n}{n} = 1$ And the $$\lim_{n \rightarrow \infty} 1 = 1$$

5. anonymous