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Math Help - Find the limit - inverse trig

  1. #1
    DBA
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    Find the limit - inverse trig

    Find the limit:

    lim x-> 0+ tan ^-1 (lnx) =?

    I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...
    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
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    lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DBA View Post
    Find the limit:

    lim x-> 0+ tan ^-1 (lnx) =?

    I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...
    Thanks
    I sure will give you another limit equation, son. If we let \ln(x)=z this becomes \lim_{z\to-\infty}\text{arctan}(z). Is the answer clear now?
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  4. #4
    Eater of Worlds
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    Let's do something wacky and fun.

    Note that tan^{-1}(u)=\frac{i}{2}\left(ln(1-ix)-ln(1+ix)\right)

    =\frac{i}{2}ln(\frac{1-ix}{1+ix})

    =\frac{i}{2}ln\left(\lim_{x\to 0}\frac{2-2i\cdot ln(x)}{1+(ln(x))^{2}}-1\right)

    Use L'hopital on the inside:

    \frac{i}{2}ln\left(\lim_{x\to 0}\frac{-i}{ln(x)}-1\right)

    The limit tends to -1 and we get \frac{i}{2}ln(-1)

    ln(-1)={\pi}i

    So, we get \frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}
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  5. #5
    DBA
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    Quote Originally Posted by Calculus26 View Post
    lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2
    is lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[limx-> 0+ (ln(x))] ?
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  6. #6
    DBA
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    Quote Originally Posted by Drexel28 View Post
    I sure will give you another limit equation, son. If we let \ln(x)=z this becomes \lim_{z\to-\infty}\text{arctan}(z). Is the answer clear now?
    Hi, thanks.
    but why lim z -> -infinity?
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  7. #7
    DBA
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    Quote Originally Posted by galactus View Post
    Let's do something wacky and fun.

    Note that tan^{-1}(u)=\frac{i}{2}\left(ln(1-ix)-ln(1+ix)\right)

    =\frac{i}{2}ln(\frac{1-ix}{1+ix})

    =\frac{i}{2}ln\left(\lim_{x\to 0}\frac{2-2i\cdot ln(x)}{1+(ln(x))^{2}}-1\right)

    Use L'hopital on the inside:

    \frac{i}{2}ln\left(\lim_{x\to 0}\frac{-i}{ln(x)}-1\right)

    The limit tends to -1 and we get \frac{i}{2}ln(-1)

    ln(-1)={\pi}i

    So, we get \frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}
    Thanks, but we did not have L'hopital yet and with the number i I don't understand a thing. Sorry
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  8. #8
    MHF Contributor Calculus26's Avatar
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    yes in general

    limf(g) = f(lim(g)) at a point if x = a if g is continuous at x = a and the limf exists at g(a)
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  9. #9
    Eater of Worlds
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    I am sorry. Perhaps I shouldn't have done that. Just having fun. You will learn L'Hopital in time. It's a handy tool.

    There are easier ways to go about it. Take note that tan^{-1}({\infty})=\frac{\pi}{2}
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