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Limits of trigonometric functions

  1. Oct 3, 2008 #1
    i'm having some trouble with evaluating limits if anyone could help me out a bit i would appreciate it. thanks in advance

    evaluate the following limits:

    Question 1:
    x -> 0 [tex]\frac{2tan^{2}x}{x^{2}}[/tex]

    my answer:
    u = x[tex]^{2}[/tex]
    as x -> 0 u-> 0

    = lim
    u->0 [tex]\frac{2sin u}{u cos u}[/tex]

    = 2 lim
    u -> 0 [tex]\frac{sin u}{u}[/tex] (lim u-> 0) [tex]\frac{1}{cos u}[/tex]

    = 2 (1) [tex]\frac{1}{cos (0)}}[/tex]
    = 2 (1)(1/1)
    = 2

    question 2:

    x->0 [tex]\frac{1- cosx}{x sinx }[/tex]

    my answer:
    x ->0 [tex]\frac{sinx}{x}[/tex] (lim x-> 0 [tex]\frac{1}{1 + cosx}[/tex]


    = 1/2

    question 3:

    x -> 0 [tex]\frac{sin 7x}{sin 4x}[/tex]

    my answer:
    [tex]\frac{7sin 7x}{4 sin 4x}[/tex]

    i'm not sure where to go from here on this one...i know the resulting limit will be equal to 7/4 but i dont know how to come up with this answer...i'm getting really confused

    question 4:

    x->0 [tex]\frac{sin(cosx)}{sec x}[/tex]

    = [tex]\frac{sin(cosx)}{1/cosx}[/tex]
    = sin(cos) x . (cos x)
    = sin(cosx[tex]^{2}[/tex])
    = sin(cos (0)[tex]^{2}[/tex])
    = 0.84

    I am really unsure of what i am doing here...i know its a lot to look at, i really appreciate the help. Thank you!
  2. jcsd
  3. Oct 3, 2008 #2


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    Homework Helper

    Hello lamerali! :smile:

    Question 2 is fine.

    In Question 1, putting u = x2 makes tan2x tan2(x2).

    You should have used L'Hôpital's rule instead (or jsut tan2 = sin2/cos2)

    In Question 3, you did use L'Hôpital's rule, but you differentiated wrong … try again! :smile:

    In Question 4, sin(cos x) . (cos x) = sin(cosx2) is completely wrong … and where ever did 0.84 come from? :confused:

    You should just have put x = 0 … no limit is involved. :wink:
  4. Oct 4, 2008 #3
    Thannnnnnnk you tiny tim!!!
    question 1, 2, and 3 are completely clear, i believe i came up with the correct answer. :D

    However i'm still unsure of question 4. Even if you plug in x = 0 into [tex]\frac{sin cosx}{sec x}[/tex] where sec x is equal to [tex]\frac{1}{cos x}[/tex] my answer comes to 0.84147, also if you graph the function there appears to be a limit. i dont know where i am going wrong in my calculations on this one! have any ideas?

    thanks again for the help! :)
  5. Oct 4, 2008 #4


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    Science Advisor
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    Hi lamerali! :smile:

    Your technique for Question 4 seems fine now.

    sin(1) = 0.84147 … you've now got the right result for the right reason! :wink:
  6. Oct 4, 2008 #5
    Great!! :D thank you tiny tim!!
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