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Exponential question

  1. Oct 15, 2011 #1
    How do you differentiate lim (2^h - 1)/h ?
    h --> 0
    I would like to know how to do it algebraically instead of picking values of x that approach 0 and plugging it in to 2^x.
  2. jcsd
  3. Oct 15, 2011 #2


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    I presume you mean how do you evaluate that limit. If you know the derivative of ex you can do it this way:

    [tex]\lim_{h\rightarrow 0}\frac{2^h-1}{h}=
    \lim_{h\rightarrow 0}\frac{e^{h\ln 2}-1}{h}=
    (\ln 2)\ \lim_{h\rightarrow 0}\frac{e^{h\ln 2}-1}{h\ln 2}[/tex]

    Not let t = h ln(2) giving

    [tex](\ln 2)\ \lim_{t\rightarrow 0}\frac{e^{t}-1}{t}[/tex]

    That gives ln(2) as the answer because the limit of the fraction is 1. You can see that either by recognizing that difference quotient as the derivative of ex at x = 0 or by applying L'Hospital's rule to it if you have had that.
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