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Doubt about two limits (short)

  1. Jan 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the limit of
    ##1): \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}##
    ##2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})##
    I am not quite sure if i can solve it the way I did, it has been to easy so there's a trick, I'm afraid

    3. The attempt at a solution
    ##1) \displaystyle \lim_{n \to +\infty}(\frac{f(a+\frac{1}{n})}{f(a)})^{\frac{1}{n}}##
    =##\displaystyle \lim_{n \to +\infty} e^{\frac{1}{n} (log(f(a+\frac{1}{n})-log(f(a)))}##=
    =##e^0=1##

    ##2) \displaystyle \lim_{x \to a} (\frac{f(x)}{f(a)})^{\frac{1}{ln(x)-ln(a)}}(=1^{\infty})##
    ##\displaystyle \lim_{x \to a} e^{ln(\frac{f(x)}{f(a)}) \frac{1}{ln(x)- ln(a)}}##
    ##\displaystyle \lim_{x \to a} \frac{ln(f(x))-ln(f(a))}{ln(x)- ln(a)}(=\frac{0}{0})## with de l'Hopital
    ##\displaystyle \lim_{x \to a} \frac{f'(x)}{xf(x)}=\frac{f'(a)}{af(a)}##
     
  2. jcsd
  3. Jan 5, 2013 #2

    SammyS

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    The derivative of ln(x) is 1/x.

    You have 1/x in the denominator. That becomes x in the numerator. (You could have checked your result with an appropriate example.)
     
  4. Jan 5, 2013 #3

    Ray Vickson

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    Note: that should be l'Hospital"---like the place with the emergency ward---not "l'Hopital" (although it is pronounced as "lo-pee-tal", with a silent 's').
     
  5. Jan 5, 2013 #4
    I tend to see it as l'Hôpital, so maybe they read l'Hôpital and didn't know how to type an ô?
     
  6. Jan 6, 2013 #5
    thanks everyone!
     
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