find the limit ....

Aug 2008
172
1
\(\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n} \)
 
Jan 2010
354
173
Since the numerator and denominator both approach infinity individually, you may use L'Hopitals rule to get:

\(\displaystyle \lim_{n \to \infty} \frac{\ln (n+1)}{\ln n} = \lim_{n \to \infty} \frac{\frac{d}{dn} \ln (n+1)}{\frac{d}{dn} \ln n} = \lim_{n \to \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{n+1}\)
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Dec 2009
1,506
434
Russia
\(\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n} \)
Using L'Hupital rule, we get:
\(\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n} = \lim_{n\rightarrow \infty } \frac{n}{n+1} =1 \)