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Math Help - solve exponential limit

  1. #1
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    solve exponential limit

    Hi everyone, there is a limit problem I am trying to solve for an hour now.
    I need to solve it without using l'H˘pital's rule because we haven't covered it in the course yet.

    I now that solution is -1, I just can't find the way.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by drugcpp View Post
    Hi everyone, there is a limit problem I am trying to solve for an hour now.
    I need to solve it without using l'H˘pital's rule because we haven't covered it in the course yet.

    I now that solution is -1, I just can't find the way.
    Observe that \left(\frac{1}{2}\right)^x=2^{-x}. So we have the following:

    \begin{aligned}\lim\limits_{x\to0}\frac{2^x-1}{2^{-x}-1} &= \lim\limits_{x\to 0}\frac{2^x(2^x-1)}{2^x(2^{-x}-1)}\\ &= \lim\limits_{x\to0}\frac{2^x(2^x-1)}{1-2^x}\\ &= \ldots\end{aligned}

    Something nice will happen when you continue simplifying this limit...
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  3. #3
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    Right,
    1-2^x = −(2^x-1) and that's all. Thank you!
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  4. #4
    MHF Contributor chisigma's Avatar
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    Is 2^{x}= e^{x\ \ln 2} and 2^{-x}= e^{-x\ \ln 2} so that...

    \displaystyle \frac{e^{x\ \ln 2}-1}{e^{-x\ \ln 2}-1}= \frac{e^{\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2}}\ \frac{e^{\frac{x}{2}\ \ln 2} - e^{-\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2} - e^{\frac{x}{2}\ \ln 2}}= - \frac{e^{\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2}}

    ... and now the task is easy...

    Kind regards

    \chi \sigma
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