help finding limit

Discussion in 'Math' started by zJakez, Oct 2, 2006.

  1. zJakez

    Thread Starter New Member

    Oct 2, 2006
    2
    0
    I don't know if I'm allow to ask calculus questions on this forum, but here goes:
    Find the limit of (e^-3x)(cos9x) when x--->infinity. State whether the limit is (some numbers), negative infinity/positive infinity, does not exist, or use any possible ways to find the limit.

    My calculations:

    Method1:

    I put it in the calculator and used the data table to replace x with any positive numbers since x--> +infinity. The answer is -infinity, since x replace with any positive numbers the result will always came out negative.

    Method2:

    Get the equation out of the limit and make it an equation: y = (e^-3x)(cos9x)
    Then I used the derivative of it, and since it multiplies I have to use the Product Rules.

    y = (-3e^-3x)(cos9x) + (e^-3x)(-9sin9x) = 3e^-3x(-cos9x - 3sin9x)

    It is not the answer but my teach would accept an equation written out with limit.

    Am I approaching this equation correctly?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Code ( (Unknown Language)):
    1.  
    2.                              cos(9x)
    3. y  = (e^(-3x))(cos(9x))  =  --------
    4.                              e^(3x)
    5.  
    I think your equation can be rewritten in the form above. If I haven't goofed something up then as x approaches positive infinity it looks like the denominator will approach infinity. That would mean that as x ---> infinity, y would approach 0. The numerator is nothing more than a sinusoidal waveform with amplitude 1 and a period of 0.22pi.

    hgmjr
     
  3. menouf2005

    New Member

    Oct 4, 2006
    1
    0
    what about take the transformation of cos(x)=(e(x)+e(-x))/2
    so

    e(9x)+e(-9x)
    y=------------------
    2*e(3x)
     
  4. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    As far as I can remember, cos(x) is not equal to (e^(x) + e^(-x))/2. The identities are as follow:

    cos(x) = (e^(ix) + e^(-ix))/2

    and

    cosh(x) = (e^(x) + e^(-x))/2
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    n9352527 is correct here.

    Consider:

    cos(jx) = cosh(x)

    We know:

    cos(x) = (e^(jx) + e^(-jx))/2

    Let x = jx:

    cos(jx) = (e^(jjx) + e^(-jjx))/2

    And j*j = -1

    Therefore:

    cos(jx) = (e^(x) + e^(-x))/2 = cosh(x)

    You can do the same for sin(x) and sinh(x), from which you can get tanh(x) = sinh(x)/cosh(x)

    Dave
     
Loading...