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Limit of a function as x goes to 1

  1. Sep 22, 2013 #1
    f(x) = 1-cos(x-1) / (x-1)2

    Can someone please explain how this can be done other than L'Hospital's rule? I tried but I just don't understand how. I know the answer is 1/2 though.

    I tried separating to 2 limits but i got ∞ - ∞ and that is obviously wrong but I don't know what my mistake was.
     
  2. jcsd
  3. Sep 22, 2013 #2
    You could do this by doing a series expansion for the numerator and looking at the limit of the leading order expression. This is a useful series expansion:

    cos(x-1) ≈ 1 - (x-1)2/2! + (x-1)4/4! + ...
    This is basically the definition of cosine, but you can get it through taylor expansion about x=1. Note that as x goes to 1, the higher order terms are negligible.
    If you plug this into the numerator of your expression, you get some nice cancellation happening.
     
  4. Sep 22, 2013 #3

    mfb

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    There are brackets missing for the numerator, I think.

    If you know good upper and lower estimates for cos() around 0, you can use them. The full taylor series is the better solution, if you can use that.
     
  5. Sep 23, 2013 #4

    Mark44

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    A Taylor's Series expansion seems to me to be the way to go.
     
  6. Sep 23, 2013 #5
    How would you go about even expressing that as a taylor series o_O
     
  7. Sep 23, 2013 #6

    Office_Shredder

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    iRaid, can you express cos(x-1) as a Taylor series around x=1?

    Actually Jolb has it above, but it doesn't do much if you don't know why that is what it is.
     
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