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Limits with trig functions

  1. Feb 4, 2004 #1
    I am having trouble with the following problems:

    1) lim as x -> 0 of (sin 3x)/2x

    2) lim as x -> 0 of (tan 5x)/(sin 2x)

    3) lim as x -> 0 of (sin²3x)/2x

    4) lim as h -> 0 of [(h+x)³ -x³]/h

    5) lim as h -> 0 of [1/(x+h) - 1/x]/h


    *Thanks for your help
     
    Last edited: Feb 4, 2004
  2. jcsd
  3. Feb 4, 2004 #2
    These are basic formulas

    for eg u can write sinx~x when x->0 similarly for tanx

    What Have u covered so far in Limits this will help us to give u better explanations
     
  4. Feb 4, 2004 #3
    We know that sin and cos are continuous,
    the lim as x-> 0 (sin (x))/x = 1
    the lim as x-> 0 (1 - cos(x))/x = 0
    and some basic trig identities
     
  5. Feb 4, 2004 #4

    HallsofIvy

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    It may be that your problem is tjat "(sin 3x)/2x" has different coefficients for the x inside and outside of the sine. That no big deal. First take out the "1/2": (1/2)(sin3x)/x and the multiply both numerator and denominator by 3: (3/2)(sin3x)/3x Think of the "3x" as u and the problem is (3/2) (sin u)/u. Since u= 3x goes to 0 when x does, what is the limit of that as u goes to 0?
     
  6. Feb 4, 2004 #5
    I got that one now, I'm really stuck on (sin² 3x)/2x
    Also, I know there is a trick for finding it in the calculator, since there is no sin² button, but I don't remember it.
     
  7. Feb 4, 2004 #6
    write it as sin3x*sin3x/2x and now find the limit it would be zero
     
  8. Feb 4, 2004 #7

    matt grime

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    1) lim as x -> 0 of (sin 3x)/2x



    set u = 3x as halls of ivy suggests



    2) lim as x -> 0 of (tan 5x)/(sin 2x)

    tan(5x) = sin(5x)/cos(5x) and it suffices to find out what sin(5x)/sin(2x) tends to so you could work it all out in terms of sin2x

    or look at the end of the post



    3) lim as x -> 0 of (sin²3x)/2x


    4) lim as h -> 0 of [(h+x)³ -x³]/h

    expand the bracket


    5) lim as h -> 0 of [1/(x+h) - 1/x]/h

    have you done basic algebra? cos that again is 'just do the manipulation' - think before asking, you'll learn a lot more


    in general you want to use l'Hopital's rule.
     
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