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Thread: limit

  1. #1
    Jan 2009


    This one has my entire class stumped:

    $\displaystyle \lim_{x\rightarrow0}\frac{\sin(\tan(x))-\tan(\sin(x))}{\arcsin(\arctan(x))-\arctan(\arcsin(x))}=$

    I've tried l'H˘pital but I still get $\displaystyle \frac{0}{0}$, and trying to build Taylor sums out of this mess seems impossible. I thought of using the squeeze theorem, but I have no idea what to squeeze it between or how to prove it. My attempt with l'H˘pital showed me it was > 0, and graphing it on my computer it looked to be about 1. Any help writing a proof will be greatly appreciated.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Quote Originally Posted by realpart1/2 View Post
    This one has my entire class stumped:

    $\displaystyle \lim_{x\rightarrow0}\frac{\sin(\tan(x))-\tan(\sin(x))}{\arcsin(\arctan(x))-\arctan(\arcsin(x))}=$
    This looks brutal. The only way I can see to do it is to expand the top and bottom as power series. To do this for the numerator, use the series
    $\displaystyle \sin x = x-\tfrac16x^3 + \tfrac1{120}x^5-\tfrac1{7!}x^7 + \ldots$ and $\displaystyle \tan x = x + \tfrac13x^3 + \tfrac2{15}x^5 + \tfrac{17}{315}x^7+\ldots$. Then
    $\displaystyle \tan(\sin x) = \bigl(x-\tfrac16x^3 + \tfrac1{120}x^5-\tfrac1{7!}x^7\bigr) + \frac13\bigl(x-\tfrac16x^3 + \tfrac1{120}x^5\bigr)^3 + \frac2{15}\bigl(x-\tfrac16x^3 \bigr)^5 + \frac{17}{315}x^7$ (plus powers higher than 7), with a similar expression for $\displaystyle \sin(\tan x)$. When you subtract these, I think that the lowest surviving power of x in $\displaystyle \tan(\sin x) - \sin(\tan x)$ will be something like $\displaystyle \tfrac1{20}x^7$.

    Now do the same thing for the denominator, using the power series for the inverse trig functions, and presumably the lowest surviving power of x will again be some multiple of $\displaystyle x^7$. I don't have the stamina to check that.
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  3. #3
    Super Member redsoxfan325's Avatar
    Feb 2009
    Swampscott, MA
    Using Maple to do the power series (because we all know how to take derivatives, and finding Taylor series for complicated functions is tedious and not particularly enriching), we have the Taylor series in the numerator as $\displaystyle -\frac{1}{30}x^7 -\frac{29}{756}x^9-\frac{1913}{75600}x^{11}-...$ The series in the denominator is $\displaystyle -\frac{1}{30}x^7 + \frac{13}{756}x^9 -\frac{2329}{75600}x^{11}+...$

    Thus we have $\displaystyle \frac{-\frac{1}{30}x^7 -\frac{29}{756}x^9-\frac{1913}{75600}x^{11}-...}{-\frac{1}{30}x^7 + \frac{13}{756}x^9 -\frac{2329}{75600}x^{11}+...}$

    Factoring out the $\displaystyle -\frac{1}{30}x^7$ and taking the limit as $\displaystyle x\to 0$ tells us that this limit equals 1.

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