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Lim X->0 Sin^-1X/X

  1. Dec 1, 2011 #1
    Sry that i might post up silly question but i am still new to calculus ....


    How do i solve this ??
    Lim X->0 Sin^-1X/X
     
  2. jcsd
  3. Dec 1, 2011 #2

    jgens

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    Re: Limits

    Off the top of my head, I can think of two ways you can evaluate the limit:

    1) You could apply L'Hopital's Rule.
    2) You could compute the first few terms in the series expansion of arcsin(x)/x.

    I cannot think of any elegant arguments that show this, or arguments that do not use techniques like series expansion of L'Hopital's Rule. Maybe someone else knows a clever argument for this.
     
  4. Dec 1, 2011 #3
    Re: Limits

    Could you substitute [itex]x = \sin \theta [/itex] since [itex] \sin \theta \to 0 [/itex] as [itex] \theta \to 0 [/itex]?
     
  5. Dec 1, 2011 #4

    jgens

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    Re: Limits

    You could write y = sin-1(x). Then sin(y) = x. So the OP would then just need to consider limy→0y/sin(y). That works nicely.
     
  6. Dec 1, 2011 #5

    Office_Shredder

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    Re: Limits

    This is a pretty tough limit to calculate. You could use fancy techniques, but this is the limit that's used to calculate the derivative of sin(x) usually, so it doesn't make sense to use the derivative of sin in order to calculate this limit. Typically this limit is calculated using geometry and the squeeze theorem: draw a right triangle on the unit circle and see if you can find any upper and lower bounds for sin(x)/x in terms of things that have more easily calculated limits
     
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