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Big O taylor series truncate

  1. Sep 25, 2011 #1
    I have to prove that
    [tex] \cos(x) = 1 - \frac{x^2}{2} + O(x^4)[/tex] [tex] (x \to 0) [/tex]

    My ugly attempt:
    [tex] \lim_{x \to 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{x^4}[/tex]

    [tex] \lim_{x \to 0} \frac{\cos(x) - 1}{x^4} + \frac{1}{2x^2}[/tex]

    [tex] \lim_{x \to 0} \frac{\sin(x)}{4x^3} + \frac{1}{2x^2}[/tex]

    [tex] \lim_{x \to 0} \frac{1}{4x^2} + \frac{1}{2x^2}[/tex]

    [tex] \lim_{x \to 0} \frac{1}{4x^2} + \frac{2}{4x^2} = \frac{3}{4x^2} = \infty [/tex] (It should be a finite number)

    Something does not work here, any help? Thanks!
  2. jcsd
  3. Sep 25, 2011 #2


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    Have you looked at the Taylor series with remainder theorem?
  4. Sep 25, 2011 #3
    Hi LCKurtz,
    I was searching an alternative (and elementary) way to prove it (otherwise there is nothing to prove, right?)

    My thought was: It should work using L'Hopital some finite numer of times, doesn't it? I don't know what am I doing wrong.

  5. Sep 25, 2011 #4


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    Yes, you can do it that way:

    [tex]\lim_{x\rightarrow 0}\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}
    =\lim_{x\rightarrow 0}\frac{-\sin(x)+x}{4x^3}[/tex]

    and keep going as long as you have 0/0 form. You get 1/24 eventually.
  6. Sep 25, 2011 #5
    You are completly right! Thanks!

    I think my mistake was in the second step:

    [tex] \lim_{x \to 0} \frac{\cos(x) - 1}{x^4} + \frac{1}{2x^2} [/tex]

    The limit of a sum is the sum of the limit only of both limits exists right? But the second one clearly is [tex] +\infty [/tex] and that isn't a real number, so that theorem does not apply, am I right now?

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