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Math Help - yet another limit...

  1. #1
    Member Jones's Avatar
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    yet another limit...

    Hi,

    I have this limit i would like some help with

     \displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{sin(2x)}

    So i tried replacing the sin(2x) term by it's equivalent taylor polynomials for (sin x and cos x)

    like so:
     \displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{2\bigg(\frac{x^3}{3}+\frac{x^5}{5}\bigg)\bigg(1-\frac{x^2}{2}+\frac{x^4}{4}\bigg)}

    But that didn't really seem to have helped =/
    any ideas?
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  2. #2
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    \lim_{x\to 0}tan^{-1}(\frac{1}{|x|})\frac{\sqrt{1+{\pi}x}-1}{sin(2x)}

    L'Hopital:

    \lim_{x\to 0}\frac{1}{4}\frac{\pi}{\sqrt{1+{\pi}x}\cdot cos(2x)}\cdot tan^{-1}(\frac{1}{|x|})

    The radical term can be eliminated because it evaluates to 1:

    \frac{\pi}{4}\frac{\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})}{\lim_{x\to 0}cos(2x)}

    The cos term in the denominator evaluates to 1 and we have:

    \frac{\pi}{4}\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})

    \lim_{x\to 0}tan^{-1}(\frac{1}{|x|})=\frac{\pi}{2}

    So, we have \frac{\pi}{4}\cdot \frac{\pi}{2}=\boxed{\frac{{\pi}^{2}}{8}}
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  3. #3
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    Quote Originally Posted by Jones View Post
    Hi,

    I have this limit i would like some help with

     \displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{sin(2x)}

    So i tried replacing the sin(2x) term by it's equivalent taylor polynomials for (sin x and cos x)

    like so:
     \displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{2\bigg(\frac{x^3}{3}+\frac{x^5}{5}\bigg)\bigg(1-\frac{x^2}{2}+\frac{x^4}{4}\bigg)}

    But that didn't really seem to have helped =/
    any ideas?

    Perhaps there's a mistake since the limit, as given, doesn't exist: the left and right limits are different, but if we look for the limit when  x\rightarrow 0^+ then:
    your idea looks just fine to me, but instead of writing the MacClaurin series for sine and cosine let us use the series for \sin 2x=2x+\frac{(2x)^3}{3!}+\ldots , and now we can try L'Hospitals rule on that fraction, since \arctan\frac{1}{|x|}\xrightarrow [x\to 0]{}\frac{\pi}{2} , so derivating numerator and denominator we get:

    \lim_{x\to 0}\frac{\displaystyle{\frac{\pi}{2\sqrt{1+\pi x}}}}{2+4x^2+\ldots}=\frac{\pi}{4} , and thus the limit is \frac{\pi^2}{8}

    Tonio
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  4. #4
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    Hello, Jones!

     \displaystyle\lim_{x\to 0}\,\arctan\!\left(\frac{1}{|x|}\right)\,\frac{\sq  rt{1+\pi x}-1}{\sin2x}
    Note that: . \arctan\left(\frac{1}{|x|}\right) \:=\:\text{arccot}|x|


    Multiply by \frac{\sqrt{1+\pi x} + 1}{\sqrt{1+\pi x} + 1}

    . . \text{arccot}|x|\cdot\frac{\sqrt{1+\pi x} - 1}{\sin2x}\cdot\frac{\sqrt{1+\pi x} + 1}{\sqrt{1+\pi x} + 1} \;=\;\text{arccot}|x|\cdot\frac{(1+\pi x) - 1}{\sin2x(\sqrt{1+\pi x} + 1}

    . . =\; \text{arccot}|x|\cdot\frac{\pi x}{\sin2x(\sqrt{1+\pi x} + 1)} \;=\;\frac{\pi}{2}\cdot\text{arccot}|x|\cdot\frac{  1}{\sqrt{1+\pi x} + 1}\cdot \frac{2x}{\sin2x}


    \lim_{x\to0}\left[\frac{\pi}{2}\cdot\text{arccot}|x| \cdot\frac{1}{\sqrt{1+\pi x} + 1} \cdot\frac{2x}{\sin2x}\right]

    . . =\; \frac{\pi}{2}\cdot \underbrace{\left(\lim_{x\to0}\text{arccot}|x|\rig  ht)}_{\frac{\pi}{2}} \cdot \underbrace{\left(\lim_{x\to0}\frac{1}{\sqrt{1+\pi x} + 1}\right)}_{\frac{1}{2}} \cdot \underbrace{\left(\lim_{x\to0}\frac{2x}{\sin2x}\ri  ght)}_{1}

    . . =\;\frac{\pi^2}{8}

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  5. #5
    Member Jones's Avatar
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    Hm, difficult limit. I don't think i would have pulled this one off

    Thanks for the help though
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