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Inverse func Limits

  1. Apr 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate
    [tex] \lim_{x\rightarrow 0} \ \frac{sin(tanx)-tan(sinx)}{sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x)} [/tex]


    2. Relevant equations


    3. The attempt at a solution

    I have no idea about evaluating inverse function limits.(denominator)
    I tried L^Hopital's rule but it the derivatives still remain in 0/0 form
    Thx for any help
     
  2. jcsd
  3. Apr 17, 2007 #2

    Gib Z

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    how many times have you tried the rule?
     
  4. Apr 17, 2007 #3
    I applied L^Hopital's once.

    BTW, by [itex]\ tan^{-1}x \ [/itex] I mean [itex] \ \ arctanx , not \frac{1}{tanx} [/itex]
     
  5. Apr 17, 2007 #4

    Gib Z

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    Yup I know what you meant with the arctan thing.

    Try using l'hopitals rule again.
     
  6. Apr 17, 2007 #5
    Here's what i have,

    [tex] D^r \ = \ sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x) [/tex]

    [tex]D^r'\ = \ \frac{1}{ \sqrt{1-tan^{-1}x}}. \frac{1}{1+x^2} - \frac{1}{(1+sin^{-1}x)^2}. \frac {1}{ \sqrt {1-x^2} } [/tex]

    Now i use this- :
    As [itex] \ x \ \rightarrow 0 , tan^{-1}x \ \rightarrow 0 [/itex]
    As [itex] \ x \ \rightarrow 0 , sin^{-1}x \ \rightarrow 0 [/itex]

    So, my denominator goes 0. :frown:
    Taking another derivative would be nasty
     
  7. Apr 17, 2007 #6
    Whoa...i just got an idea.
    Could i try replacing [itex] \ sin^{-1}x [/itex] by [itex] \ tanx [/itex] and so on as all tend to 0 with x tending to 0 ?
     
  8. Apr 17, 2007 #7

    HallsofIvy

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    It doesn't follow that [itex]sin^{-1} x[/itex] and [itex] tan x[/itex] tend to 0 at the same rate and that is the crucial thing.
     
  9. Apr 17, 2007 #8

    Gib Z

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    You didn't differentiate properly.

    The derivative for the bottom should be
    [tex]\frac{1}{1+x^2} \frac{1}{\sqrt{1-(\tan^{-1}x)^2}} - \frac{1}{\sqrt{1-x^2}} \frac{1}{1+(\sin^{-1}x)^2}[/tex].

    Either way, the denominator still goes to zero. Sure its nasty, but just do it. If you really can't be bothered, type it into here: www.calc101.com. It'll find the derivative for you, but entering it is just as much of a hassle as find the derivative yourself.
     
  10. Apr 18, 2007 #9
    Just forgot to put the square...thx for bringing it into notice.
    I havent seen limits with inverse trig functions in them . And now i cant even try substitutions as Halls said . If any1 figures a way out, do hint it here :biggrin:
     
  11. Apr 18, 2007 #10

    Gib Z

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    I already told you! Use L'hopitals again! It wont be pretty but bad luck.
     
  12. Apr 22, 2007 #11
    Finally found the solution -:
    LINK
     
  13. Apr 22, 2007 #12

    Gib Z

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    Thats quite a nice solution. Really he just truncated the taylor series and computed it that way, so its not such a new method, but none of us thought of that so good work.
     
  14. Apr 22, 2007 #13

    matt grime

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    L'Hopital is trucating Taylor series.
     
  15. Apr 22, 2007 #14

    Gib Z

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    Could you provide more info on that, I dont quite follow...In the link the solver truncated after 4 terms, so lhopital wud work after 4 tries?
     
  16. Apr 22, 2007 #15

    matt grime

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    If you want the limit of f(x)/g(x) as x tends to zero (say). Take the Taylor series of each about 0. You can read off the limit. That is what L'Hopital's rule is: looking at the leading coeffecients of Taylor series. What else do you think you're doing when differentiating and plugging in other than finding the coefficients of a Taylor series?
     
  17. Apr 22, 2007 #16
    [​IMG]
    I dont understand whats o(x^7) mean ? Could some1 explain
     
  18. Apr 22, 2007 #17

    Gib Z

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    Its sort of an error term. Really the series for sin and tan have an infinite number of terms, but he just just terms up to x^7, then o(x^7), or terms less important that x^7. When he says less important than, he means the error is less than. So if x is 1, the approximation we got from the first 4 can not be more than 1 off. This is what I THINK, learning from Hurkyl's posts in my thread in Calc (Not HW section).

    Thanks Matt for the reply, i understand now.
     
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