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Limit as x goes to infinity of e^(-x)* sin(x)

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data
    I am trying to take the following limit


    lim as x goes to infinity of ( e^-x )*sin(x)

    2. Relevant equations



    3. The attempt at a solution

    Can I say that it ges to '0' just because the 1/e^x goes to '0'. Or there is a better way to solve it?
     
  2. jcsd
  3. Jan 22, 2010 #2

    vela

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    Re: Limit

    It depends on how rigorous you need to be, but it goes to 0 essentially for the reason you said.
     
  4. Jan 22, 2010 #3
    Re: Limit

    Is there any other reason why it goes to zero?
     
  5. Jan 22, 2010 #4
    Re: Limit

    you can't compute a limit of a product based solely on the limit of one of the factors. use the squeeze theorem.
     
  6. Jan 22, 2010 #5

    vela

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    Re: Limit

    No. It's just that if this problem is for a math class and you're covering limits, you might be expected to justify your answer better.
     
  7. Jan 22, 2010 #6
    Re: Limit

    so if i split the 2 function apart....

    the e^-x approach '0'

    but the sin(x) approach nothing it will keep going from -1 to 1

    right?
     
  8. Jan 22, 2010 #7

    vela

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    Re: Limit

    Yes.
     
  9. Jan 22, 2010 #8

    Mark44

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    Re: Limit

    I think that the best approach is one that ice109 suggested earlier - the squeeze theorem. Note that e-x = 1/ex.

    For all real x, -1 <= sin(x) <= 1
    so, also for all real x, -1/ex <= sin(x)/ex <= 1/ex

    The leftmost and rightmost expressions approach zero as x approaches infinity, squeezing the expression in the middle. Its limit is likewise zero.
     
  10. Jan 22, 2010 #9
    Re: Limit

    i have no clue what you're talking about. what you stated isn't simply unrigorous - it's completely wrong

    if f(x)->0 that doesn't imply that f(x)g(x)->0.

    take f(x)= 1/x and g(x)=sin(x) or f(x)=x and g(x) = exp(1/x)

    for what you said to be true both limits have to exist and be finite
     
  11. Jan 23, 2010 #10

    vela

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    Re: Limit

    That's not what I said. For this particular problem, the exponential does cause the limit to be zero. It's just an example of the typical tendency of a decaying exponential factor to drive a function to a limit zero.
     
  12. Jan 23, 2010 #11
    Re: Limit

    that is what you implied.

    but fine - what is the limit of f(x)g(x) when x->0 if

    f(x) = e^-x and g(x)= exp(1/x) ?
     
  13. Jan 24, 2010 #12

    vela

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    Re: Limit

    No, it's what you inferred.

    I knew you would respond like this and therefore questioned whether I should bother responding to you at all in the first place. Yes, I could and should have been clearer, but I'm not really looking to get into a pissing contest with you about what I meant by what I wrote.
     
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