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Evaluate math limit question

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data

    use lim as x__>0 (sin(x))/(x) to evaluate the following : lim (x^(2) cot(x) + sin(x/2) - cos(x) + 1)/(x) as x__>0




    3. The attempt at a solution
    1- lim as x__>0 (x^(2) cot(x))/ (x)= x cotx = x (cosx)/(sinx)= (x/sinx) cosx = 1.1=1
    And I think I can use L'Hospital's Rule to evaluate the previous one, right?
    2- lim as x__>0 (sin(x/2))/(x)= 1/2 (sin(x/2))/(x/2)= 1/2.
    3. this is where it gets hard. I tried to use L'Hospital's Rule, but I just want to see if that makes sense or not. lim as x__>0 (-cosx +1)/(x)= 0/0. using L'H, I will get lim as x__>0 sinx= 0 . so my final answer will be 1+1/2+-0= 1.5. Is this right?
     
  2. jcsd
  3. May 18, 2010 #2

    Mark44

    Staff: Mentor

    Re: Evaluate

    Right answer, but you need to carry the limit symbol all the way until you actually take the limit. For example, in 1, you should say lim (x^(2) cot(x))/ (x)= lim x cotx = lim x (cosx)/(sinx)= lim (x/sinx) * lim cosx = 1 * 1 = 1. (All limits are as x --> 0.)

    Many instructors will ding you if you write something such as (x/sinx) cosx = 1.1=1, which isn't generally true.
     
  4. May 18, 2010 #3
    Re: Evaluate

    Thanks for pointing out the limit symbol.
    And again thanks for checking my answer.
     
  5. May 18, 2010 #4

    Cyosis

    User Avatar
    Homework Helper

    Re: Evaluate

    It strikes me as odd that you were allowed to use L'hopitals since the exercise states 'use lim as x__>0 (sin(x))/(x) to evaluate the following'. If you could use L'hopitals why even bother with giving you that limit?
     
  6. May 18, 2010 #5

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi philippe311! :smile:

    (write "->0" … it's easier to read … and try using the X2 and X2 tags just above the Reply box :wink:)
    3. is a bit difficult to read … if you're saying limx->0 (1-cosx)/x = limx->0 sinx/1 = 0, then 1 2 and 3 are all correct. :smile:

    However, the question doesn't ask you to use l'Hôpital's rule except specifically for sinx/x … perhaps you should use one of the standard trigonometric identities … 1 - cosx = 2sin2(x/2)) ? :wink:
     
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