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Limit with integral

  1. Mar 29, 2009 #1
    Limit with integrel

    1. The problem statement, all variables and given/known data
    equationrender.png

    2. Relevant equations



    3. The attempt at a solution
    I tried to take
    x=ln(u)
    dx=du/u
    and solve the integral but I keep getting stuck at the int(cos(u)/u.)
    Can anyone help me out here?
     
  2. jcsd
  3. Mar 29, 2009 #2

    Dick

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    Re: Limit with integrel

    You can't take the limit x->infinity. x is a dummy integration variable. You must mean n->infinity. Right?
     
  4. Mar 29, 2009 #3
    Re: Limit with integrel

    I meant N.
    Sorry, its a force of habit.
     
  5. Mar 29, 2009 #4

    Dick

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    Re: Limit with integrel

    Ok, then as n->infinity then 1/n -> 0. You don't actually have to do the integral to know what the limit is, because cos(e^x) is bounded. What's the limit of the integral of a bounded function between 0 and 1/n as n->infinity?
     
  6. Mar 29, 2009 #5
    Re: Limit with integrel

    so the integral would be zero as n approaches inf. since the limits of integration go from zero to zero, right?
     
  7. Mar 29, 2009 #6

    Dick

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    Re: Limit with integrel

    No, it's not just the limits of integration, you have to think about how the function you are integrating behaves. -1<=cos(e^x)<=1. Agree? So what limits can you make for the integral from 0 to 1/n of cos(e^x)? What happens as n->infinity?
     
  8. Mar 29, 2009 #7
    Re: Limit with integrel

    I'd try the MVT.
     
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