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Math Help - limit

  1. #1
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    limit

    find the limit of the function:

    lim {ln(1+x)/(3^x -1)}
    x-> 0
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  2. #2
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    Quote Originally Posted by the prince View Post
    find the limit of the function:

    lim {ln(1+x)/(3^x -1)}
    x-> 0
     \lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}
     = \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}
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  3. #3
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    why its \lim_{x \to 0} \frac{ \ln (1+x)}{x} = 1 and how did you find  \lim_{x \to 0} \frac{(3^x -1)}{x}= \log_{e}{3}

    can you give me a little advice please?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ramiee2010 View Post
     \lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}
     = \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}
    This is useless without an explanation of what you think you are doing.

    Also

    \lim_{x \to 0} \frac{ \ln (1+x)}{x}=1

    needs to be justified as does:

    \lim_{x \to 0} \frac{(3^x -1)}{x}=\ln(3)

    Hence you may as well just have suggested the OP goes straight to L'Hopital's rule to find the limit.

    And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    This is useless without an explanation of what you think you are doing.

    Also

    \lim_{x \to 0} \frac{ \ln (1+x)}{x}=1

    needs to be justified as does:

    \lim_{x \to 0} \frac{(3^x -1)}{x}=\ln(3)

    Hence you may as well just have suggested the OP goes straight to L'Hopital's rule to find the limit.



    CB
    these are identities which are taught us in 12th standard
    \lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right )
     =\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right  ) =1

    and
     \lim_{x \to 0} \frac{(a^x -1)}{x} = \lim_{x \to 0} \left ( \frac {1+x log_{e}{a}+ \frac {x^2}{2!} (log_{e}{a})^2+.... -1}{x} \right )

     = \lim_{x \to 0} \left ( log_{e}{a}+ \frac {x}{2!} (log_{e}{a})^2+.... \right ) = log_{e}{a}
    And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.
    O.K.. Captain
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ramiee2010 View Post
    these are identities which are taught us in 12th standard
    \lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right )
     =\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right  ) =1

    and
     \lim_{x \to 0} \frac{(a^x -1)}{x} = \lim_{x \to 0} \left ( \frac {1+x log_{e}{a}+ \frac {x^2}{2!} (log_{e}{a})^2+.... -1}{x} \right )

     = \lim_{x \to 0} \left ( log_{e}{a}+ \frac {x}{2!} (log_{e}{a})^2+.... \right ) = log_{e}{a}
    O.K.. Captain
    Well they are not universally taught (because its a waste of brain space to remember things that can be rederived in 10 seconds) and so they need to be justified.

    CB
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