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Trigonometric limit question

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Lim (1-cosX)/X^2

    2. Relevant equations
    Not 100% sure try to use the following identity:
    sin^2(X)=(1-cosX)(1+cosX) or

    3. The attempt at a solution
    Tried substituting (sin^2(X))/(1+cosX)) for 1-cosX but that didnt help and so for the last half hour i've been staring at this problem and I am still lost. Somebody help me out.
    Possible answers were 0, .5, 1, 2
  2. jcsd
  3. Oct 13, 2008 #2


    Staff: Mentor

    There are at least a couple of ways to do this, but one of them might not be applicable based on which class you're in.

    Try multiplying the numerator and denominator by (1 + cos(x))/(1 + cos(x)). Then make your substitution. If you do that, the limit, as x --> 0, will be
    lim [tex]\frac{sin^2(x)}{x^2(1 + cos(x))}[/tex]
    Now, do you know any limits that involve sin(x) and x, as x --> 0?
  4. Oct 13, 2008 #3
    Im in AP calculus B/C if it affects your answer at all
    and if i'm not mistaken

    limit sin(x)/x = 1
    x-> 0


    limit sin^2(x)/x^2 = 1
    x-> 0

    but how would i go about seperating the the 1+cos(x) from the denominator

    could i make the function

    limit (sin^2(x)/x^2)*(1/(1+cosx))
    x-> 0

    then substitute in for x?
    Last edited: Oct 13, 2008
  5. Oct 13, 2008 #4


    Staff: Mentor

    Not quite, but a property of limits is that lim (A * B) = lim A * lim B, providing that all the limits exist. Is that enough of a hint?

    The other approach is to look at the infinite series for 1 - cos(x). I hesitated to point you in that direction since I didn't know where you were in your course. If you're interested in the details of this approach, let me know.

  6. Oct 14, 2008 #5


    User Avatar
    Science Advisor

    Yes, you certainly can write
    [tex]\frac{1- cos(x)}{x^2}= \frac{(1- cos(x))(1+ cos(x))}{x^2(1+ cos(x))}[/tex]

    [tex]= \frac{1-cos^2(x)}{x^2(1+ cos(x))}= \frac{sin^2(x)}{x^2(1+ cos(x))}[/tex]

    [tex]= \left(\frac{sin(x)}{x}\right)^2 \frac{1}{1+ cos(x)}[/tex]

    That should make the limit easy.
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