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Math Help - Know one (limit)

  1. #1
    Math Engineering Student
    Krizalid's Avatar
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    Known one (limit)

    \forall a>0, prove that \lim_{x\to0}\frac{a^x-1}x=\ln a

    I like this limit, 'cause most of people the first thing they do is apply L'H˘pital's Rule.
    Last edited by Krizalid; December 1st 2007 at 04:24 AM. Reason: Removing title
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    \forall a>0, prove that \lim_{x\to0}\frac{a^x-1}x=\ln a

    I like this limit, 'cause most of people the first thing they do is apply L'H˘pital's Rule.
    I (almost) never used L'Hopital's Rule.

    a^x = e^{x\ln a} = 1+(x\ln a) + \frac{(x\ln a)^2}{2!}+...
    Thus,
    \frac{a^x-1}{x} = \ln a + \frac{x\ln^2 a}{2!}+\frac{x^2\ln a}{3!}+....
    Since power series are uniformly convergent we can pass the limit through to get,
    \ln a + 0 + 0 +... = \ln a.


    ---
    Solution #2.

    This is just the derivative of a^x at 0.
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  3. #3
    MHF Contributor red_dog's Avatar
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    a^x-1=u\Rightarrow x=\frac{\ln(1+u)}{\ln a}
    Then \displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{u\to 0}\frac{u\ln a}{\ln(1+u)}=
    \displaystyle =\ln a\lim_{u\to 0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\displaystyle\ln\lim_{u\to 0}(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\ln e}=\ln a
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    a^x-1=u\Rightarrow x=\frac{\ln(1+u)}{\ln a}
    Then \displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{u\to 0}\frac{u\ln a}{\ln(1+u)}=
    \displaystyle =\ln a\lim_{u\to 0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\displaystyle\ln\lim_{u\to 0}(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\ln e}=\ln a
    nice solution!

    i believe this is the first time i've seen someone bring a limit inside a log. i didn't know you could do that
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  5. #5
    MHF Contributor
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    Quote Originally Posted by Jhevon View Post
    i believe this is the first time i've seen someone bring a limit inside a log. i didn't know you could do that
    The reason you can do it is that the log function is continuous. That means that if x\to a then \ln x\to \ln a.
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  6. #6
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    Okay

    Here's another solution similar to red_dog's one

    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
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