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Proof that lim x->0 log(1+x)/x = 1

  1. Jan 7, 2008 #1
    Given that [tex]\log{(1+x)}=\int_0^x\frac{dt}{1+t}[/tex], how would one prove that [tex]\lim_{x \to 0}\frac{\log{(1+x)}}{x}=1[/tex]?
     
  2. jcsd
  3. Jan 7, 2008 #2
    Just use l'Hôpital's rule and the fundamental theorem of calculus.
     
  4. Jan 7, 2008 #3
    Thanks! I had thought of that. However, L'Hôpital's rule is introduced much later in this book so I assume I'm not supposed to use it. It's so far covered logarithms, differentiation and integration techniques... any other ideas?
     
  5. Jan 7, 2008 #4
    Ok, if you can't use l'Hôpital, how about this? Subtract a log(1) from the numerator in the expression of the limit. It should look quite familiar now. :-D
     
  6. Jan 7, 2008 #5
    Sorry, I should have mentioned something: You're supposed to prove it using the identity I gave in the first part of my post. This is part B of an exercise in apostol. Part A was to prove it using your method, which I did.

    The statement of part A was something like "Prove the relationship using the definition of f'(1), where f(x) = log(x)" or something of that nature.
     
  7. Jan 7, 2008 #6

    D H

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    Express [itex]1/(1+t)[/itex] as an infinite series, integrate from 0 to x, divide by x, and take the limit as [itex]x\to 0[/itex].
     
  8. Jan 7, 2008 #7
    Meh. The book hasn't covered series' either. Oh well.
     
  9. Jan 8, 2008 #8

    Gib Z

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    I haven't tried this yet, normally I would go with either of Monchot's suggestions, but you could try expressing the integral in terms of Riemann sums. I think that should work.
     
  10. Jan 8, 2008 #9
    For the expression

    [tex]I=\frac{1}{x}\,\int_0^x\frac{d\,t}{1+t}[/tex]

    apply the change of variables [itex]t=x\,u,d\,t=x\,du[/itex], so

    [tex]I=\int_0^1\frac{d\,u}{1+x\,u}[/tex]

    Then

    [tex]\lim_{x \rightarrow 0}I=\int_0^1 \lim_{x \rightarrow 0}\left(\frac{1}{1+x\,u}\right)\,d\,u=\int_0^1 d\,u=1[/tex].

    You are allowed to change order between the integral and the limit, since the integral exists by hypothesis.
     
  11. Jan 8, 2008 #10

    Gib Z

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    Rainbow Child's suggestion crushes mine :( But I just tried it, and it works as well =]
     
  12. Jan 8, 2008 #11
    Thank you Rainbow Child! I understand your proof and it makes perfect sense. Very clever!
     
  13. Sep 17, 2011 #12
    Lim x->0 (x(1+(x/2)+((x^2)/3)+.................))/x
    =lim x->0 (1+(x/2)+((x^2)/3)+.................)
    =1+0+0+0......................=1
     
  14. Sep 17, 2011 #13

    HallsofIvy

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    As lovely as it is, Rainbow Child's suggestion requires knowing that, in this case,
    [tex]\lim_{x\to 0} \int_0^1 \frac{dt}{1+ xt}= \int_0^1 \lim_{x\to 0}\frac{dt}{1+ xt}[/tex]
    which, while true, probably requires deeper math than uman, who has not yet dealt with L'Hopital's rule or power series, has available.
     
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