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Math Help - help with two limits

  1. #1
    Member Jones's Avatar
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    help with two limits

    Hi

    I have two limits i would like some help with

    first one:

    \lim \limits_{x \to 0}\frac{10^x-e^x}{x}

    and second one

    \lim \limits_{x \to \infty} x(2tan^{-1}x-\pi)


    The first limit i thought could be rewritten like this [Math]\frac{e^{x*ln10}-e^x}{x}[/tex]
    which would then be simplified down to:

    \frac{10e^x-e^x}{x}

    and then using L'Hopital rule to aquire 9

    The last one i don't know how to start
    Last edited by Jones; December 28th 2009 at 10:17 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) \lim_{x\to 0}\frac{e^x\left(\left(\frac{10}{e}\right)^x-1\right)}{x}=

    =\lim_{x\to 0}e^x\cdot\lim_{x\to 0}\frac{\left(\frac{10}{e}\right)^x-1}{x}=1\cdot ln\frac{10}{e}

    because \lim_{x\to 0}\frac{a^x-1}{x}=\ln a

    2) \lim_{x\to\infty}\frac{2\arctan x-\pi}{\frac{1}{x}}

    and use l'Hospital.
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  3. #3
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    How you prove that this term exist ?

     \lim_{x\to 0}\frac{a^x-1}{x}=\ln a
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by gilyos View Post
    How you prove that this term exist ?

     \lim_{x\to 0}\frac{a^x-1}{x}=\ln a
    It is well known that \lim_{x\to 0}(1+x)^{\frac{1}{x}}=e

    Then \lim_{x\to 0}\ln(1+x)^{\frac{1}{x}}=\lim_{x\to 0}\frac{\ln(1+x)}{x}=1.

    Now let a^x-1=y\Rightarrow x=\log_a(1+y) and y\to 0

    \lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\lim_{y\to 0}\frac{y}{\ln(1+y)}\cdot\ln a=

    =\lim_{y\to 0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}\cdot\ln a=\ln a
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Going off of red dog's initial response we see that \lim_{x\to\infty}\frac{2\arctan(x)-\pi}{\frac{1}{x}} may be transformed by letting z=\frac{1}{x} into \lim_{z\to0}\frac{2\arctan\left(\tfrac{1}{z}\right  )-\pi}{z} but \arctan\left(\tfrac{1}{z}\right)=\frac{\pi}{2}-\arctan(z) were it readily follows that our limit is equal to -2\lim_{z\to0}\frac{\arctan(z)}{z} where it follows from the definition of the derivative that our answer is -2\cdot\frac{1}{1+0^2}=-2.
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  6. #6
    Member Jones's Avatar
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    Hm, sorry i didn't follow everything.

    How did you know to divide (2~arctan ~x -\pi) by 1/x
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jones View Post
    Hm, sorry i didn't follow everything.

    How did you know to divide (2~arctan ~x -\pi) by 1/x
    I followed what red dog suggested
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