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Another difficult limit

  1. Dec 23, 2012 #1
    1. The problem statement, all variables and given/known data
    here it is:
    the teacher suggeste to solve it with de l'hopital, however it is not necessary

    ## \displaystyle \lim x \to +\infty\ (\frac{a^x -1}{x(a-1)})^\frac{1}{x}##
    with ## a>0## , ##a≠1##

    3. The attempt at a solution
    the indeterminate formula is ## (\frac{\infty}{\infty})^0 ##
    but i thought of changing the variable ##x \to \infty## to ##z \to 0##
    such that i have the known:
    ## \frac{(a^{\frac{1}{t}}-1)}{\frac{1}{t}} = log(a)##
    so i obtain:
    ##(\frac{log(a)}{a-1})^{+\infty} ##
    but i don't know how to go on..
     
  2. jcsd
  3. Dec 23, 2012 #2
    Why don't you simply use the fact that the limiting operator respects division when the limits on the numerator and the denominator are defined? It simplifies the problem to a great extent.
     
  4. Dec 23, 2012 #3
    ok, then it should be:

    ##\displaystyle \lim z \to 0 (\frac{a^{(1/z)-1}}{1/z})^z * \displaystyle \lim z \to 0 (\frac{1}{a-1})^z##
    ##=(log(a))^0*(\frac{1}{a-1})^0##
    =1

    is it correct?
     
  5. Dec 23, 2012 #4

    SammyS

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    Your post #3 result is incorrect.

    If [itex]\displaystyle \lim \, \ln(f(x))=L\,,\ \text{ then }\ \lim\, f(x)=e^L\ . [/itex]

    So, look at [itex]\displaystyle \lim_{x \to +\infty\ }\ \ln\left(\left(\frac{a^x -1}{x(a-1)}\right)^{1/x}\right)\ .[/itex]
     
  6. Dec 31, 2012 #5
    I've found a possible solution, but I'm a bit dubious, could you please check it?
    1st situation: $0<a<1$

    ##\displaystyle \lim_{x \to +\infty} (\frac{a^x-1}{x(a-1)})^{\frac{1}{x}}=\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}##

    If ## \lim log(f(x)) = L## then ##\lim f(x) = e^L##

    ##=\frac{1}{+\infty} log(\frac{-1}{-\infty})(=\frac{-\infty}{+\infty})##

    With de l'Hopital

    ##\stackrel{\text{H}}{=} \frac{x(a-1)}{a^x-1} \frac{x a^x log(a) (a-1) - (a-1) (a^x-1)}{x^2 (a-1)^2}$ = $\frac{x a^x log(a) - (a^x-1)}{(a^x-1)x}##

    = ##\frac{a^xloga}{a^x-1}-\frac{1}{x}##

    ##= \frac{0}{-1}-0=0##

    ## e^0 = 1##

    Situation 2: $a>1$

    ##\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}=##
    ## =e^{ 0* log (\frac{\infty}{\infty})}##

    I use de l'Hopital within the log (am I allowed to do this?)

    ##\displaystyle \lim_{x \to +\infty} \frac{1}{x}log(\frac{a^x loga}{a-1})## = ##\frac{\infty}{\infty}##

    de l'Hospital again

    ##\displaystyle \lim_{x \to +\infty} \frac{a-1}{a^x loga} \frac{a^xlog^2a+0-0}{(a-1)^2}$ = $\frac{loga}{a-1}##

    ##=e^{\frac{loga}{a-1}}##

    I didn't want to use the known limit = ##\frac{a^x-1}{x}=loga## because x had to go to 0 to apply that, while here it doesn't.
     
    Last edited: Dec 31, 2012
  7. Dec 31, 2012 #6

    Dick

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    No, don't try the "l'Hopital within the log". You are making life a lot harder by not using rules of logs. Split your log up into log(a^x-1)-log(x)-log(a-1).
     
  8. Dec 31, 2012 #7
    oh, allright, i didn't think about it. so now it should be:

    ## \lim_{x \to +\infty}e^{\frac{1}{x} (log(a^x-1)-log(x)-log(a-1))}## = ##e^{\frac{\infty}{\infty}}##

    using de l'Hopital i get:

    ##\frac{1}{a^x-1}log(a)-\frac{1}{x}-\frac{1}{a-1}## diving both numerator and denominator of the first member it becomes

    ##\frac{log(a)}{1-\frac{1}{a^x}}-\frac{1}{x}-\frac{1}{a-1}##=##log(a)-\frac{1}{a-1}##

    =##\frac{e^{log(a)}}{e^{\frac{1}{a-1}}}##

    Is everything ok now?
    at the beginning was it good to divide the limit in two different cases?
    Thanks a lot again
     
  9. Dec 31, 2012 #8

    Dick

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    Yes, it's a very good idea to divide it into different cases. a^x behaves differently in the two cases. You are getting there. The derivative of log(a-1) is zero isn't it? It's a constant! And you forgot a factor of a^x in the first term, but the next line is correct, so I guess that's a typo.
     
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