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Proving equation involving limits without derivatives

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    This is not really a homework or a coursework question. But I got a warning that I should submit my post in this section of the website.. I'm just saying this because I don't know if the answer to my question is at all achievable. And if it is how I should go about trying to solve it. Anyway.. Here is my question:

    Can I prove that:
    [tex] \lim_{h \to 0} \frac{(\frac{a^h-1}{h})}{(\frac{b^h-1}{h})} = \lim_{h \to 0} \frac{a^h-1}{b^h-1} = log_ba [/tex]
    without using derivatives?

    Of course the first equality is trivial But I thought it might be important, that's why it is there.

    2. Relevant equations

    Honestly don't know..

    Maybe the squeeze theorem?

    3. The attempt at a solution

    I tried making both sides an exponent of b. Which would looks like:
    [tex] b^{log_ba} = b^{\frac{a^h-1}{b^h-1}} [/tex]
    which means:
    [tex] a = b^{\frac{a^h-1}{b^h-1}} [/tex]
    I don't even know if that is the right path though..
     
  2. jcsd
  3. Mar 1, 2013 #2

    Dick

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    You could use the taylor series of a^h=exp(log(a)*h) and similarly for b. But, of course, the taylor series involves derivatives. I'm not really sure you can even define what a^h even means without using derivatives in some way. I'm not sure you should bother trying.
     
  4. Mar 2, 2013 #3

    lurflurf

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    Derivatives are inherent to that, though they could be hidden.


    $$\lim_{h \rightarrow 0} \frac{a^h-1}{h}=\log(a)$$

    can be derived a number of ways and is analogous to

    $$\lim_{h \rightarrow 0} (1+a h)^h=e^a$$
     
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