# find the limit ....

#### flower3

$$\displaystyle \lim_{n\rightarrow \infty } \frac{ln(n+1)}{ln \ n}$$

#### drumist

Since the numerator and denominator both approach infinity individually, you may use L'Hopitals rule to get:

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

#### chisigma

MHF Hall of Honor
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### Also sprach Zarathustra

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