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Math Help - limits / equations?

  1. #1
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    limits / equations?

    for what value of a is the following equation true?

    lim ((x+a)/(x-a))^x = e
    x->inf
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  2. #2
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    Re: limits / equations?

    Take the natural log of both sides, so that you wind up with:

    \lim_{x\to\infty}\frac{\ln\left(\frac{x+a}{x-a} \right)}{\frac{1}{x}}=1

    Now use L'H˘pital's rule to find a.
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  3. #3
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    Re: limits / equations?

    after applying L'H˘pital's rule i got the limit as x approaches infinity ((x-a)(-2a))/(-x^-2) = 1
    and now what do i do from there?
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  4. #4
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    Re: limits / equations?

    You have differentiated the numerator incorrectly. You want:

    \frac{d}{dx}\left(\ln\left(\frac{x+a}{x-a} \right) \right)=\frac{1}{\frac{x+a}{x-a}}\cdot\frac{-2a}{(x-a)^2}=\frac{2a}{a^2-x^2}

    Now, putting this together with the derivative of the denominator, we have:

    \lim_{x\to\infty}\frac{2ax^2}{x^2-a^2}=1

    which we can write as:

    2a\lim_{x\to\infty}\frac{1}{1-\frac{a^2}{x^2}}=1

    From here, it is a piece of cake.
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  5. #5
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    Re: limits / equations?

    im not seeing how this is a piece of cake?
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    Re: limits / equations?

    Quote Originally Posted by pnfuller View Post
    im not seeing how this is a piece of cake?
    LEARN this: \lim _{x \to \infty } \left( {1 + \frac{a}{{x + b}}} \right)^{cx}  = e^{ac}

    Yours can be rewritten an \left( {1 + \frac{{2a}}{{x - a}}} \right)^x .
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: limits / equations?

    Sorry, bad choice of words.

    What is \lim_{x\to\infty}\frac{a^2}{x^2} ?
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    Re: limits / equations?

    the limit as a^2/x^2 approaches infinity equals 0? is that what you mean?
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    MHF Contributor MarkFL's Avatar
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    Re: limits / equations?

    Yes, so what does that tell you about the limit at the end of post #4?
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  10. #10
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    Re: limits / equations?

    that the limit as x approaches infinity is 1/1 and it equals 1 but what does a equal?
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    Re: limits / equations?

    Reread that equation carefully. You were told \displaystyle \begin{align*} 2a \cdot \lim_{x \to \infty}\frac{1}{1 - \frac{a^2}{x^2}} = 1 \end{align*}
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: limits / equations?

    Okay, since the limit is 1, you simply have:

    2a=1

    Now solve for a.
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  13. #13
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    Re: limits / equations?

    so a = 1/2
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    Re: limits / equations?

    Quote Originally Posted by pnfuller View Post
    that the limit as x approaches infinity is 1/1 and it equals 1 but what does a equal?
    This whole thread is totally frustrating for me.
    Please look at reply #5.
    The answer is a=0.5 . WHY?
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  15. #15
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    Re: limits / equations?

    why is it frustrating for you?
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