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Math Help - find the limit ....

  1. #1
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    find the limit ....

     \lim_{n\rightarrow \infty }  \frac{ln(n+1)}{ln \ n}
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  2. #2
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    Since the numerator and denominator both approach infinity individually, you may use L'Hopitals rule to get:

    \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}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

    \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 n \rightarrow \infty of (1)...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by flower3 View Post
     \lim_{n\rightarrow \infty }  \frac{ln(n+1)}{ln \ n}
    Using L'Hupital rule, we get:
     \lim_{n\rightarrow \infty }  \frac{ln(n+1)}{ln \ n} =  \lim_{n\rightarrow \infty } \frac{n}{n+1} =1
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