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Math Help - constant to the power of x???

  1. #1
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    constant to the power of x???

    so here's the problem....

    lim [ (2^x - 1) / x ] as x-> 0

    i never get these right. how would you do this?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by runner07 View Post
    so here's the problem....

    lim [ (2^x - 1) / x ] as x-> 0

    i never get these right. how would you do this?
    are you allowed to use L'hospital's Rule?
    use it..
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  3. #3
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    yes i know. but how do you derive 2^x. thats my main problem.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by runner07 View Post
    yes i know. but how do you derive 2^x. thats my main problem.

    ah, ok... D_x [a^x] = a^x \ln a..
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  5. #5
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    okay. thanks !
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by runner07 View Post
    lim [ (2^x - 1) / x ] as x-> 0
    More generally:

    Let a>0.

    \lim_{x\to0}\frac{a^x-1}x=\ln a.

    Let's set a change of variables according to u=\frac1{a^x-1}\implies x=\log_a\left(1+\frac1u\right)\,\therefore\,u\to\i  nfty, this yields

     <br />
\begin{aligned}<br />
\mathop {\lim }\limits_{x \to 0} \frac{{a^x - 1}}<br />
{x} &= \mathop {\lim }\limits_{u \to \infty } \frac{1}<br />
{{u \cdot \log _a \left( {1 + \dfrac{1}<br />
{u}} \right)}}\\<br />
&= \mathop {\lim }\limits_{u \to \infty } \frac{1}<br />
{{\log _a \left( {1 + \dfrac{1}<br />
{u}} \right)^u }}.<br />
\end{aligned}<br />

    Which finally yields

    \mathop {\lim }\limits_{u \to \infty } \frac{1}<br />
{{\dfrac{{\ln \left( {1 + \dfrac{1}<br />
{u}} \right)^u }}<br />
{{\ln a}}}} = \mathop {\lim }\limits_{u \to \infty } \frac{{\ln a}}<br />
{{\ln \left( {1 + \dfrac{1}<br />
{u}} \right)^u }} = \ln a\,\blacksquare
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  7. #7
    Eater of Worlds
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    The Big K always has a nice approach. Here's another.

    \lim_{x\rightarrow{0}}\frac{2^{x}-1}{x}

    Let u=2^{x}-1, \;\ x=\frac{ln(u+1)}{ln(2)}

    Upon making the subs:

    ln(2)\lim_{u\rightarrow{0}}\frac{u}{ln(u+1)}

    Now, using the first few terms of the Taylor expansion for ln(u+1)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}....... and simplifying, we get:

    ln(2)\lim_{u\rightarrow{0}}\frac{6}{2u^{2}-3u+6}

    ln(2)(1)=ln(2)
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  8. #8
    Senior Member DivideBy0's Avatar
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    Wow, how do you guys know to do all these wacky substitutions? Are these tricks you've memorized? Or like, do you think 'Let's make a sub in there and see what happens next', or can you see it all straight away? For instance with Krizalid's sub I would never have seen it coming in a hundred years
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Wow, how do you guys know to do all these wacky substitutions? Are these tricks you've memorized? Or like, do you think 'Let's make a sub in there and see what happens next', or can you see it all straight away? For instance with Krizalid's sub I would never have seen it coming in a hundred years
    Kriz typically comes up with some fantastic ways of looking at this stuff. For the rest of us (not usually me when it comes to limits) it's a function of having done many of them before and a bit of "I've seen something similar to this before and this substitution worked..."

    -Dan
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  10. #10
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    Quote Originally Posted by DivideBy0 View Post
    Wow, how do you guys know to do all these wacky substitutions? Are these tricks you've memorized? Or like, do you think 'Let's make a sub in there and see what happens next', or can you see it all straight away? For instance with Krizalid's sub I would never have seen it coming in a hundred years
    It comes from experience


    Note: You can just realize that \lim (a^x - 1)/x is the derivative of a^x at zero. So why do all this fancy stuff?
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  11. #11
    Eater of Worlds
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    So why do all this fancy stuff?
    Because it's fun and we can. Occam's Razor be damned.
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