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Simple Trig Limit

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    lim x->0 [tex](1-cosx)/sinx[/tex]

    2. Relevant equations
    [tex](sinx)/(1+cosx)[/tex] = [tex](1-cosx)/(sinx)[/tex]

    [tex]sinx/x[/tex] = 1

    3. The attempt at a solution
    Well, I know I need to change my original limit, as I cannot let the denominator be 0. I usually try to factor out things, but there is nothing to factor out here. I'm not really sure what to do.
  2. jcsd
  3. Sep 19, 2010 #2
    L'Hopital's rule might work well here. Have a look at that.
  4. Sep 19, 2010 #3
    Im afraid I don't know how to use this. I am in Calculus AB, so it is not something we learn. Any other ideas?

    We usually factor, rationalize, etc. to find limits.
  5. Sep 19, 2010 #4

    [itex] \frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x} [/itex]

    then, won't the limits of the two equations be equal?
  6. Sep 19, 2010 #5
    L'Hoptial's rule is extremely simple, and it makes solving the problem very obvious in this case. Anyways, with the equation you gave, the two limits will be the same as Strants said. You don't need any fancy work to evaluate that limit if you use that equation, because you can evaluate it exactly at the point x=0.
  7. Sep 19, 2010 #6
    I would think so, but when you graph this, it looks like there could be many limits.
  8. Sep 19, 2010 #7

    ok, thanks!
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