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Limits involving natural exponential2

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    lim(e-2x cosx)
    x-> infinity

    3. The attempt at a solution

    I thought right away that the limit would not exist because of cos x oscillating the function between + and - but the answer in the book says zero. I need help figuring out why my thinking was incorrect.
     
  2. jcsd
  3. Nov 3, 2009 #2

    Dick

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    -1<=cos(x)<=1. Whether there's a limit depends on what e^(-2x) does. What does it do? For a proof think about using the squeeze theorem.
     
  4. Nov 3, 2009 #3

    lanedance

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    think about what happens to the negative exponential as x gets large, and maybe try a squeeze theorem
     
  5. Nov 3, 2009 #4
    e^(-2x) will get very small.
     
  6. Nov 3, 2009 #5
    well it will still be oscillating but approaching zero all the same I suppose
     
  7. Nov 3, 2009 #6

    Dick

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    Ok, so what's you conclusion? And why?
     
  8. Nov 3, 2009 #7

    Dick

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    That's fine, if you don't have to prove it and omit the 'I suppose'.
     
  9. Nov 3, 2009 #8
    How would I apply the squeeze theorem to this problem?
     
  10. Nov 3, 2009 #9

    Dick

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    Can you find two functions f(x) and g(x) such that f(x)<=cos(x)e^(-2x)<=g(x) such that f(x) and g(x) both approach zero? Possibly using -1<=cos(x)<=1?
     
  11. Nov 3, 2009 #10
    Because
    -1<=cos(x)<=1

    -e-2x <= cos(x)e-2x <= e-2x

    is that it? I just emulated what I have seen on a few math help sites. But it makes sense now that I have thought it out a bit.

    soo

    lim -e-2x=0=lim e -2x
    x->infinity x->infinity

    so

    lim cos(x)e-2x=0
    x->infinity
     
  12. Nov 3, 2009 #11

    Dick

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    Brilliant. A little emulation goes a long way.
     
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