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Prove that arctanx has to be less than x

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that :
    $$\lim_{x\to 0}\frac{tan^{-1}x}{x}<1$$
    And
    $$\lim_{x\to 0}\frac{sin^{-1}x}{x}>1$$
    2. Relevant equations

    None
    3. The attempt at a solution

    I have to prove that arctanx has to be lesser than x.
    It's derivative is 1 at x=0 and keeps decreasing as x increases. So it's slope decreases from 1.
    Also, at x=0, arctanx=x. So the curve which initially touched y=x goes below it. So arctanx<x.
    Is this correct?
     
  2. jcsd
  3. Mar 31, 2015 #2

    PeroK

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    Why not think about tan(x) and sin(x)?
     
  4. Mar 31, 2015 #3
    I know that sinx<x.
    So x<arcsinx..... By taking arcsin on LHs and RHS.
    Thank you.
    Is my method in post 0 correct?
     
  5. Mar 31, 2015 #4

    PeroK

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    It might be neater to consider the functions

    f(x) = x - arcsin(x)

    g(x) = x - arctan(x)

    Shouldn't the limits be ##\le## and ##\ge##?

    Why not use L'hopital to calculate them?
     
  6. Mar 31, 2015 #5

    vela

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    You can't prove these because they're false. The limit is a single number, i.e.,
    $$\lim_{x \to 0}\frac{\tan^{-1}x}{x} = 1.$$
     
  7. Mar 31, 2015 #6

    Ray Vickson

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    The statements you are trying to show are both false. If you think ##\lim_{x \to 0} \arctan(x)/x < 1##, how much < 1 do you think it is? Is it < 0.9? Is it < 0.99?

    Also: arctan(x) is not < x; in fact:
    [tex] \arctan(x) \begin{cases} < x,& x > 0\\ > x, & x < 0
    \end{cases} [/tex]
    Just plot the two functions to see that this is true.

    You should likewise check ##\sin(x)## against ##x## for small ##x## of both signs, + and -.
     
  8. Mar 31, 2015 #7
    sorry... its x->0+
     
  9. Mar 31, 2015 #8
    so you can prove it for x->0+ by using LH rule directly and substituting 0+.
     
  10. Apr 1, 2015 #9

    SammyS

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    Look at Post #5 from vela .

    The limits you are referring to are all exactly equal to zero one .

    (Thanks to RGV !)
     
    Last edited: Apr 1, 2015
  11. Apr 1, 2015 #10

    Ray Vickson

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    Actually, exactly equal to one.
     
  12. Apr 1, 2015 #11
    No. its x->0+. Using graph, the arctanx curve starts coming below y=x.
     
  13. Apr 1, 2015 #12

    SammyS

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    Do you understand limits ?
     
  14. Apr 1, 2015 #13
    yes. its a homework question to prove that $$\lim_{x\to 0}\sqrt{\frac{tan^{-1}x}{x}-\frac{sin^{-1}x}{x}}$$ does not exist and its given in the solution that ##\frac{tan^{-1}x}{x}<1## while ##\frac{sin^{-1}x}{x}>1##, when x->0.
    If it is wrong, even then ##\frac{tan^{-1}x}{x}<\frac{sin^{-1}x}{x}## when x->0 (actually 0+).
     
  15. Apr 1, 2015 #14

    SammyS

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    That's quite different from the original post.

    For 0 < x ≤ 1 , ## \displaystyle\ \frac{\tan^{-1}x}{x}-\frac{\sin^{-1}x}{x} <0##

    Taking the square root gives an imaginary result.
     
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