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Homework Help: Limits of trig functions

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    lim as x approaches 2

    lim as x approaches pi/4

    2. Relevant equations

    equations above

    3. The attempt at a solution
    for the first limit, i tried substituting t = (pi/2)-(pi/x) but i got stuck

    i have no idea how to do the second limit
  2. jcsd
  3. Oct 21, 2009 #2


    Staff: Mentor

    Are you allowed to use L'Hopital's Rule?
  4. Oct 21, 2009 #3
    no, i can only evaluate the limit as it is,

    i was thinking of using t = pi/2 - pi/x and substituting the pi/x so i have an identity, but i don't think that will work

    i have no idea on the second one, only changing tanx to sinx/cosx
  5. Oct 21, 2009 #4


    Staff: Mentor

    Well, phooey.
    I don't have any ideas for the first one, but I believe the limit is pi/4 (using Excel to compute a few values for t near 2).
  6. Oct 21, 2009 #5
    i was thinking of evaluating the limit of t at 2, so the answer would be zero, then substitute t in for pi/x and take the limit as t approaches 0

    does this help?
  7. Oct 21, 2009 #6


    Staff: Mentor

    I don't understand what you're saying. Can you elaborate in the context of your problem?
  8. Oct 21, 2009 #7
    For the first one, let u=[itex]\pi[/itex]/x. Then use the identity cos u = sin([itex]\pi[/itex]/2 - u) and rewrite the denominator so you have [itex]\pi[/itex]/2 - u in it. Use another substitution with [itex]\pi[/itex]/2 - u and you should get something like sinx/x. Don't forget to change the limits after a substitution.
  9. Oct 21, 2009 #8
    well after saying t = pi/2 - pi/x, i would then evaluate the limit of t at x=2 and get 0, since pi/2 - pi/2 = 0

    then, i would substitute the pi/x in the cos and put pi/2 - t in the cos function to get

    (cos(pi/2 - t))/(x-2), then i would take the limit as t approaches 0 since the lim as x approaches 2 of t = 0. one problem is the x in the denominator, so then i would have to solve for x in t = pi/2 - pi/x

    seeing that cos(pi/2 - t) is an identity, i would substitute that for the identity itself and solve but i can't get an answer

    i also thought that a common identity is cos(pi/2 - theta) = sin theta, so i also tried to apply that to cos(pi/2 - t) to get sin(t), but i also don't get an answer
  10. Oct 21, 2009 #9


    Staff: Mentor

    This is what I was having trouble with. You don't take the limit of t at x=2. You take the limit of t, as t approaches something.
  11. Oct 21, 2009 #10
    yeah i got my wording mixed up

    bohrok, could you elaborate on your method a little more

  12. Oct 21, 2009 #11
    limx→2 [itex]\pi[/itex]/x = [itex]\pi[/itex]/2, so use lim as u→[itex]\pi[/itex]/2 in the new limit after the substitution.
    [tex]\lim_{u\rightarrow \pi/2}\frac{\cos u}{\frac{\pi}{u} - 2} = \lim_{u\rightarrow \pi/2}\frac{\sin(\frac{\pi}{2} - u)}{\frac{\pi}{u} - 2} = \lim_{u\rightarrow \pi/2}\frac{\sin(\frac{\pi}{2} - u)} {\frac{2}{u}(\frac{\pi}{2} - u)}[/tex]

    Then use t = [itex]\pi[/itex]/2 - u so you can finally use the limit sin(t)/t
  13. Oct 21, 2009 #12
    The second one is involved, at least the way I did it.

    First I wrote tanx as sinx/cosx and wrote the numerator as one fraction. The part I worked with that's making it indeterminate was (cosx-sinx)/(x-[itex]\pi[/itex]/4)
    Multiply by the conjugate and use the identity with cos2x, let t=2x, then use the identity cosθ = sin([itex]\pi[/itex]/2 - θ). Use another substitution so you get a limit with sinx/x.
  14. Oct 22, 2009 #13
    Hi Bohrok,

    I'm just evaluating this limit for fun. Could you please reveal the last substitution which you made to get [tex] \frac{\sin x}{x} [/tex]?

  15. Oct 22, 2009 #14
    Once you get to the equation with [itex]\cos(2x)[/itex], you can multiply top and bottom by 2 to get the desired form. There may be another substitution that would do the same thing.
  16. Oct 22, 2009 #15
    [tex]\lim_{x\rightarrow \pi/4} \frac{\tan x - 1}{x - \frac{\pi}{4}} = \lim_{x\rightarrow \pi/4} \frac{\frac{\sin x}{\cos x} - \frac{\cos x}{\cos x}}{x - \frac{\pi}{4}} = \lim_{x\rightarrow \pi/4} \frac{\cos x - \sin x}{x - \frac{\pi}{4}} \times \frac{-1}{\cos x}[/tex]

    [tex]\frac{\cos x - \sin x}{x - \frac{\pi}{4}} \times \frac{\cos x + \sin x}{\cos x + \sin x} = \frac{\cos^2x - \sin^2x}{(x - \frac{\pi}{4})(\cos x + \sin x)} = \frac{\cos2x}{x - \frac{\pi}{4}} \times \frac{1}{\cos x + \sin x}[/tex]

    Let u = 2x and x = u/2. As x→[itex]\pi[/itex]/4, u→[itex]\pi[/itex]/2

    [tex]\lim_{x\rightarrow \pi/4} \frac{\cos2x}{x - \frac{\pi}{4}} = \lim_{u\rightarrow \pi/2} \frac{\cos u}{\frac{u}{2} - \frac{\pi}{4}} = \lim_{u\rightarrow \pi/2} \frac{\sin(\frac{\pi}{2} - u)}{(-\frac{1}{2})(\frac{\pi}{2} - u)}[/tex]

    Let t = [itex]\pi[/itex]/2 - u. As u→[itex]\pi[/itex]/2, t→0

    [tex]\lim_{u\rightarrow \pi/2} \frac{\sin(\frac{\pi}{2} - u)}{\frac{\pi}{2} - u} = \lim_{x\rightarrow 0} \frac{\sin t}{t}[/tex]

    And that takes care of the 0/0 part of the limit. Put everything together and you can find the limit.
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