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Homework Help: Finding right limit of log(2sin(x/2))-log(2(sin(x/2)+cos(x/2))

  1. Aug 1, 2010 #1
    log(2sin(x/2))-log(2(sin(x/2)+cos(x/2))

    I have to find the right limit from x --> 0

    How can I solve this limit?
     
    Last edited: Aug 1, 2010
  2. jcsd
  3. Aug 1, 2010 #2

    Mentallic

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    Re: Limit

    Have you tried anything? How about simplifying the two logarithms and then using double angle formulae on the trigs.
     
  4. Aug 1, 2010 #3
    Re: Limit

    If you simplify them you get:

    log(sin(x/2)/(sin(x/2)+cos(x/2))

    But if I fill in 0 than I get log(0) but that doesn't exist..
    If I use a double angle formulae I also get log(0)
     
  5. Aug 1, 2010 #4

    Mentallic

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    Re: Limit

    Well that's why it's a limit which suggests you need to figure out what the limit is as x approaches 0 from the right side. Where is it heading?
     
  6. Aug 1, 2010 #5
    Re: Limit

    If I plot the function on my calculator, I thinks it's heading to -8, but I'm not sure
     
  7. Aug 1, 2010 #6

    Mentallic

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    Re: Limit

    That's because your calculator isn't accurate enough or you can't see exactly what's happening really close (even closer than that) to 0.

    You've already figured out that the answer is ln(0+) - the plus is there to show it's from the right side, not both sides of 0. 0- denotes from the left side.
    What is the answer to ln(0+)? That's the same as asking what happens to the function y=ln(x) as x approaches 0 from the right?
     
  8. Aug 2, 2010 #7
    Re: Limit

    Is it -∞?
     
  9. Aug 2, 2010 #8

    Mentallic

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    Re: Limit

    Yep! :smile:

    To convince yourself, try every smaller positive values of x,

    x=0.1
    x=0.001
    x=10-10

    You'll see where it's headed.
     
  10. Aug 2, 2010 #9
    Re: Limit

    Thanks!
     
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