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Evaluating limit as h->0 of (e^h-1)/h

  1. Jul 24, 2012 #1
    I'm trying to differentiate e^x from first principles but I can't find a way to manipulate this expression [tex]\frac{e^h-1}{h}[/tex] so I can evaluate the limit without getting 0/0
     
  2. jcsd
  3. Jul 24, 2012 #2

    eumyang

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    Use the definition of e:
    [itex]lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e[/itex]
    So for small values of h,
    [itex]e \approx \left( 1 + h \right)^{1/h}[/itex], or
    [itex]e^h \approx 1 + h[/itex].

    Replace eh in [itex]\frac{e^h-1}{h}[/itex] with 1 + h and go on from there.
     
  4. Jul 24, 2012 #3
  5. Jul 24, 2012 #4

    Mute

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    In order to use L'Hopital's rule, you need to know the derivative of ex, but that's what the OP is trying to prove. To use L'Hopital's rule in this case would be circular logic.
     
  6. Jul 24, 2012 #5

    eumyang

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    I'm not sure how. The OP is trying to "differentiate ex from first principles," which I assume means that he/she want to find the derivative of ex by using the limit definition:
    [tex]\frac{d}{dx}e^x = lim_{h \rightarrow 0} \frac{e^{x+h} - e^x}{h} = ...[/tex]
    I don't think you can use L'Hopitals' Rule when we "don't know" the derivative of ex yet.


    EDIT: Beaten to it by Mute. :wink:
     
  7. Jul 24, 2012 #6
    Oops my bad. I didn't read.
     
  8. Jul 25, 2012 #7
    Don't we get 0/0 again?

    Limit h->0

    [tex]\frac{1+h-1}{h}[/tex]
     
  9. Jul 25, 2012 #8

    eumyang

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    Um, you need to simplify the expression.
    [tex]lim_{h \rightarrow 0}\frac{1+h-1}{h} = lim_{h \rightarrow 0} \frac{h}{h} = ...[/tex]
     
  10. Jul 25, 2012 #9
    ^ Ah, thanks a lot...must've had a brain meltdown :p
     
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