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Limit proof

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I have to prove that the limit as n ⇒ ∞ of: n(a1/n-1) = log(a) -> For every a >0

    2. Relevant equations
    I have no idea what to use

    3. The attempt at a solution
    This was an exam question i left it blank, because i had (and have) no idea on where to even start, i've tried using integrals/antiderivatives, but i didnt seem to get anywhere, this part of the exam was about derivatives, antiderivateves, integrals and fundamental calculus theorems (which we use to plot functions, so i dont see this one applying here)
    I know that if i set "a" as 1 that checks out, but that doesnt prove anything...
    ive thought about using the left side of the equation as an exponent to "e" and setting it as equal to "a" but i dont see that helping.
    What am i supposed to use to prove this?
     
  2. jcsd
  3. Jan 29, 2016 #2

    Mark44

    Staff: Mentor

    I would start by writing ##n(a^{1/n} - 1)## as ##\frac{a^{1/n} - 1}{1/n}## and then takiing the limit.
     
  4. Jan 29, 2016 #3
    Hmm, i dont see it ...
    as n becomes greater the nth root of a goes really near to 1, and if you take 1 out of it, you get a value between 0 and 1, divided by an (more or less) equally small number, it just leaves me at the same place as not doing that.

    Should i take a derivative of both sides? that doesnt seem to help me either
    i think i dont know which "limit solving algorithm" im supposed to use
     
  5. Jan 29, 2016 #4

    Mark44

    Staff: Mentor

    So as ##n \to \infty##, the numerator approaches 0, right? What does the denominator do?
     
  6. Jan 29, 2016 #5
    also aproaches 0, because nth root goes to 1.
    im not sure how to evaluate how fast they aproach it and what the relation between it
     
  7. Jan 29, 2016 #6

    Mark44

    Staff: Mentor

    OK, so both numerator and denominator are approaching zero. There should be some rule (that's a hint) you've learned about that you can use.
     
  8. Jan 29, 2016 #7
    Riiiiight, i can only think of the l'hopital or cauchy's rule, i dont think thats it, but i'm gonna think about that for a while, and try these and other methods, thanks, i'l upate here in a bit
     
  9. Jan 29, 2016 #8

    Mark44

    Staff: Mentor

    I was hinting at L'Hopital's Rule. Why do you think it doesn't apply here?
     
  10. Jan 29, 2016 #9
    because that was one of the first things i thought of, it didnt seem like it was gonna help as getting and exponential of((1-n)/n) didnt seem like a great simplification, im trying this out now
     
  11. Jan 29, 2016 #10
    so numerator look like 1/n * a(1-n/n) and 1-n/n goes to -1.. so numerator = 1/n*a, this doesnt seem right ..

    About the denominator, I should derivate in respect to n? wether i do this or not, the result is not correct.

    am i supposed to derivate the nominator in respect to n?, i guess i should since its the unknown
     
  12. Jan 29, 2016 #11
    Ok i got it right now.
    Thanks for all the help :)
     
  13. Jan 29, 2016 #12

    Mark44

    Staff: Mentor

    Correct, that's not right. ##a^{1/n}## is an exponential function, not a power function like ##x^n##.
    "Derivate" is a word, but it's not one that's used in mathematics. You differentiate a function to get its derivative.
    Good deal. I'll mark this thread as Solved.
     
  14. Jan 29, 2016 #13
    Hahahah sorry for the bad english xD, i meant differentiate.
     
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