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Limit involving exponential function question

  1. May 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the limit as x tends to zero of: (e^-x - cos x)/2x


    2. Relevant equations
    lim_x->0 e^-x = 1
    lim_x->0 cos x = 1
    lim_x->0 sin x / x= 1


    3. The attempt at a solution
    Hi everyone,
    Here's what I've done so far:

    (e^-x - cosx)/2x = [(e^-x)^2 - (cosx)^2] / 2x(e^-x - cosx) .... multiplying by conjugate

    = [e^-2x - 1 + (sinx)^2 ] / 2x(e^-x - cosx)
    ... And so I want to try and isolate the sinx to put it over the x, which will then go to 1. But I don't know how to do this. Or am I going about it the wrong way entirely?

    Thanks for any help
     
  2. jcsd
  3. May 11, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    I don't think that is going to work. You could taylor expand both the exponential and the cosine. This allows you to find the limit pretty easily.
     
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