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Thread: L'Hopital lim (x->0) (a^x-1)/x

  1. #1
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    L'Hopital lim (x->0) (a^x-1)/x

    Good morning,

    I am doing a quick review and going over some of my completed problems here. I have the following:

    $\displaystyle
    \displaystyle{\frac{\lim}{x\rightarrow0}\frac{a^x-1}{x}}
    $

    Applying $\displaystyle L'H$ I got

    $\displaystyle
    \displaystyle{\frac{\lim}{x\rightarrow0}\frac{\ln( a)a^x}{1}}
    $

    Here's the problem, I don't remember why I went from $\displaystyle a^x$to $\displaystyle ln(a)a^x$.

    It's something simple, right?
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  2. #2
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    $\displaystyle a^x = e^{\ln(a^x)}=e^{\ln(a)x}$
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  3. #3
    MHF Contributor matheagle's Avatar
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    you differentiated $\displaystyle a^x$ which is $\displaystyle a^x\ln a$

    just like $\displaystyle {d\over dx}e^x=e^x\ln e=e^x$
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