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Calculating a limit

  1. Jun 12, 2010 #1
    I know I should apply L'Hopital's rule and use a^b=e^(b*ln(a)) but I can't finish the calculations.

    limit as x->0 ((arcsin(x))/x) ^(1/x^2)
     
  2. jcsd
  3. Jun 12, 2010 #2

    lanedance

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    thats a tricky one, so going with what you said
    [tex] \lim_{x \to 0} (\frac{arcsin(x)}{x})^{\frac{1}{x^2}}
    = \lim_{x \to 0} e^{\frac{ln(\frac{arcsin(x)}{x})}{x^2}}[/tex]

    now let
    [tex] b =\frac{ln(\frac{arcsin(x)}{x})}{x^2}[/tex]

    if the limit exists, its equal to e^(b), so finding the limit of a is sufficient

    thats 0/0 indeterminate, so we can apply L'Hops rule - though i can see it will be a bit messy
     
    Last edited: Jun 13, 2010
  4. Jun 12, 2010 #3

    lanedance

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  5. Jun 12, 2010 #4
    sorry, still a no-go. can't get the algebra together. can someone please help? i've applied L'hopital's rule 3 times and it keeps getting uglier.
     
  6. Jun 12, 2010 #5
    I think it might be helpful considering Taylor Polynomials approximation.
     
  7. Jun 12, 2010 #6
    We didn't learn yet the Taylor thingie. This assignment is about L'Hopital's rule.
     
  8. Jun 13, 2010 #7

    lanedance

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    ok, so how about starting by looking at the arcsin function and its derivatives, lets abuse the notation a bit and call it a for brevity recognising its a function of x:
    [tex]a(x) = arcsin(x), \ \ \ \ \ \ \lim_{x \to 0} a(x) = 0 [/tex]
    [tex]a'(x) = (1-x^2)^{1/2}, \ \ \ \ \lim_{x \to 0} a'(x) = 1 [/tex]
    [tex]a''(x) = x(1-x^2)^{3/2}, \ \ \ \lim_{x \to 0} a''(x) = 0 [/tex]
    [tex]a'''(x) = (1-x^2)^{3/2} -3x(1-x^2)^{5/2}, \ \lim_{x \to 0} a'''(x) = 0 [/tex]
     
    Last edited: Jun 13, 2010
  9. Jun 13, 2010 #8

    lanedance

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    now going back to
    [tex] b = \lim_{x \to 0}\frac{ln(\frac{arcsin(x)}{x})}{x^2}
    = \lim_{x \to 0} \frac{ln(a) - ln(x)}{x^2}
    [/tex]

    this is 0/0 so using L'Hop
    [tex]
    = \lim_{x \to 0} \frac{a'/a - 1/x}{x^2} = \lim_{x \to 0}\frac{1}{2} \frac{a'x - a}{ax^2}
    [/tex]

    once again, this is 0/0 so using L'Hop
    [tex]
    = \lim_{x \to 0}\frac{1}{2} \frac{a''x}{a'x^2+ 2ax}= \lim_{x \to 0}\frac{1}{2} \frac{a''}{a'x+ 2a}
    [/tex]

    one more time, this is 0/0 so using L'Hop
    [tex]
    = \lim_{x \to 0}\frac{1}{2} \frac{a'''}{a''x+a'+ 2a'}= \lim_{x \to 0}\frac{1}{2} \frac{a'''}{a''x+3a'}
    [/tex]

    and at this point you should be able to sub in with the properties of the derivatives
     
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