1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

{Calculus 1} Find the limit

  1. Aug 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The limit
    ##\lim_{x\to\pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x^{2}}##
    Can be expressed as a fraction. Solve

    2. Relevant equation

    3. The attempt at a solution


    EDIT
    See new post for solution
     
    Last edited: Aug 10, 2015
  2. jcsd
  3. Aug 9, 2015 #2
    Keep in mind, when taking the derivative, that cos(pi/2) is a constant.
     
  4. Aug 9, 2015 #3
    I know I'm getting 0 as the final answer thats not right though it should be a fraction of two relatively prime positive integers.
     
  5. Aug 9, 2015 #4
    My mistake I typed the question wrong
     
  6. Aug 9, 2015 #5
    The derivative of the top, when derived correctly, will yield a prime number. Just derive xcos(pi/x) as you would derive an x with a constant coefficient. For the bottom, pi^2 is also a constant and should go to 0 when derived. All you have left to deal with would be the -x^2.
     
  7. Aug 9, 2015 #6
    sorry it should be cos(x/2) I read it wrong
     
  8. Aug 9, 2015 #7

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Why is zero incorrect?
     
  9. Aug 9, 2015 #8

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Then that makes more sense, what do you get now?
     
  10. Aug 9, 2015 #9
    I'm sorry I typed it wrong trying again with Let t=x−π. Then x=t+π,
     
  11. Aug 9, 2015 #10
    I did not get zero as an answer. A place where you could be making a computational mistake would be in the derivation of xcos(x/2).
     
  12. Aug 10, 2015 #11

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Can you go ahead and retype the whole thing? The original limit was zero, the second correction gave a finite value, but retype everything.
     
  13. Aug 10, 2015 #12
    That would be much appreciated (:
     
  14. Aug 10, 2015 #13

    Let ##t=x-\pi## Then ##x=t+\pi##

    ##\lim_{x\to \pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x{^2}}##


    ##=\lim_{t\to 0}\frac{(\pi+t)cos\frac{(\pi+t)}{2}}{\pi^{2}-(\pi+t){^2}}##


    ##=\lim_{t\to 0}\frac{(\pi+t)sin\frac{t}{2}}{t^{2}+2\pi t}##

    ##=\lim_{t\to 0}\left(\frac{\pi + t}{2(t+2\pi)}\cdot \frac{sin\frac{t}{2}}{\frac{t}{2}}\right)##


    ##=\frac{1}{4}##


    Latex thing is neat will use in future
     
    Last edited: Aug 10, 2015
  15. Aug 10, 2015 #14
    Can't you use L'hospital's rule since you have an indeterminate limit?

    Where lim x→π = f(x)/g(x) is also = lim x→π f'(x)/g'(x)

    So just take the derivative of the top divided by the derivative of the bottom. Don't for get chain rule and that π2 is a constant, do not change π2 into 2π. A lot of calculus students make that mistake.

    Also remember the derivative of cosθ is -sinθ (A lot of calculus students forget the negative as well)
     
  16. Aug 10, 2015 #15

    Student100

    User Avatar
    Education Advisor
    Gold Member

    Can you? Sure. His solution is fine though.

    ##\lim_{x\to \pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x{^2}}##

    =

    ##\lim_{x\to \pi}\frac{xsin\frac{x}{2}}{4x}##

    =

    ##\frac{1}{4}##
     
    Last edited: Aug 10, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: {Calculus 1} Find the limit
  1. Calculus 1: Limit (Replies: 4)

Loading...