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anonymous

  • 5 years ago

Please help me ... limit to infinity of ln (n+1) / ln n

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    did you understand it? :)

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

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Thanks for your help :)

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