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Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0

  1. Nov 17, 2011 #1

    sharks

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    Gold Member

    1. The problem statement, all variables and given/known data
    Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0


    2. Relevant equations
    Maybe L'Hopital's rule.


    3. The attempt at a solution
    First, i put in zero, and got the indeterminate form 0/0
    So, i used L'Hopital's rule and got this:

    [cosx.sin^-1(x) + sinx.-sin^-2(x).cosx]/2x

    And again when putting zero, i get [1/0 + 0]/0 = infinity/0
    So i tried to differentiate again, but it's become so big, i don't know what's what in my copybook. Help!
     
  2. jcsd
  3. Nov 17, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    Write out the function as

    [tex]\frac{sin(x)}{x}\frac{sin^{-1}(x)}{x}[/tex].

    You know the limit of the first fraction. Apply L'Hospital to the second one.

    ehild
     
  4. Nov 17, 2011 #3

    HallsofIvy

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    "infinity/0" is NOT an indeterminate form! That would tell you the limit is infinity.

    However, you seem to be thinking that [tex]sin^{-1}(x)[/tex] means [tex]1/sin(x)[/tex]. That's clearly NOT the case here since, if it were, "[itex]sin(x)sin^{-1}(x)[/itex]" would just be 1 and [itex]sin(x)sin^{-1}(x)/x^2[/itex] would be just [itex]1/x^2[/itex] which does go to infinity.

    However, here, [itex]sin^{-1}(x)[/itex] is the arcsine, the inverse function to sine. The derivative of arcsin(x) is [itex]1/\sqrt{1- x^2}[/itex].

     
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