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Math Help - L'Hospital

  1. #16
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Can you explain why?
    Explain the implication? Or explain why you may interchange the limit and the natural log.

    I think you mean the impication

    \ln(y)=-1

    So exponentiating both sides gives

    e^{\ln(y)}=y=e^{-1}=\frac{1}{e}
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  2. #17
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Are you sure Maclaurin discovered Cramer's rule...I thought it was some monk...
    My trusty book tells me that Maclaurin the the original discoverer of Cramer's Rule...

    ...which monk?
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  3. #18
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    My trusty book tells me that Maclaurin the the original discoverer of Cramer's Rule...

    ...which monk?
    I don't have that book anymore ...Cramer?
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  4. #19
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    Quote Originally Posted by Mathstud28 View Post
    Explain the implication? Or explain why you may interchange the limit and the natural log.

    I think you mean the impication

    \ln(y)=-1

    So exponentiating both sides gives

    e^{\ln(y)}=y=e^{-1}=\frac{1}{e}

    Clearly. Thank you
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  5. #20
    GAMMA Mathematics
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    Quote Originally Posted by Mathstud28 View Post
    I don't have that book anymore ...Cramer?
    Cramer's Rule is the method to solve for the discriminant of a non 2x2 matrix.
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  6. #21
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by colby2152 View Post
    Cramer's Rule is the method to solve for the discriminant of a non 2x2 matrix.
    Yeah I know...I was saying "...Cramer?" Meaning that is probably who invented the method.
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  7. #22
    Eater of Worlds
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    We could perhaps try it without the Hospital rule.

    Let t=\frac{1}{1-x}, \;\ x=1-\frac{1}{t}

    Then we get the recognizable and fairly famous limit which can be solved by a suitable substitution:

    \lim_{t\to {\infty}}\left(1-\frac{1}{t}\right)^{t}=\frac{1}{e}

    Probably wasn't worth posting, but I had to post at least one thing today
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  8. #23
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    We could perhaps try it without the Hospital rule.

    Let t=\frac{1}{1-x}, \;\ x=1-\frac{1}{t}

    Then we get the recognizable and fairly famous limit which can be solved by a suitable substitution:

    \lim_{t\to {\infty}}\left(1-\frac{1}{t}\right)^{t}=\frac{1}{e}

    Probably wasn't worth posting, but I had to post at least one thing today
    Trust me, I really wanted to not use L'hopital's but...the heading man..the heading.
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