# find the limit ....

• May 28th 2010, 07:42 AM
flower3
find the limit ....
$\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n}$
• May 28th 2010, 07:53 AM
drumist
Since the numerator and denominator both approach infinity individually, you may use L'Hopitals rule to get:

$\displaystyle \lim_{n \to \infty} \frac{\ln (n+1)}{\ln n} = \lim_{n \to \infty} \frac{\frac{d}{dn} \ln (n+1)}{\frac{d}{dn} \ln n} = \lim_{n \to \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{n+1}$
• May 28th 2010, 07:57 AM
chisigma
Is...

$\displaystyle \frac{\ln (n+1)}{\ln n} = \frac{\ln (1+\frac{1}{n}) + \ln n}{\ln n} = 1 + \frac{\ln (1+\frac{1}{n})}{\ln n}$ (1)

... and now You have to valuate the limit for $\displaystyle n \rightarrow \infty$ of (1)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 28th 2010, 07:58 AM
Also sprach Zarathustra
Quote:

Originally Posted by flower3
$\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n}$

Using L'Hupital rule, we get:
$\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n} = \lim_{n\rightarrow \infty } \frac{n}{n+1} =1$