1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the limit of ln (sinx) as x approaches pi

  1. Jan 27, 2013 #1
    I need to find the limit of. Ln (sinx) as x approaches pi-, that is, as x approaches pi from the left.

    I thought I should use the squeeze theorem, but I am not sure how to apply it. My teacher is not requiring us to use the delta/Epsilon method, so I am sure it is a squeeze method. But how do I set it up?

    I tried -1 < sinx < 1,
    Thus ln (-1) < ln (sinx) < ln (1)

    But you can't take the ln of a negative... please help.
     
  2. jcsd
  3. Jan 27, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What can you say about [itex]\sin(x)[/itex] as [itex]x \rightarrow \pi^-[/itex]? Is it positive or negative? What value does it approach?
     
  4. Jan 27, 2013 #3
    As x aproaches pi from the left your sine function aproaches 0. This means this is equivalent of finding the limit as the thing inside the natural log aproaches 0. Its very easy limit.
     
    Last edited: Jan 27, 2013
  5. Jan 27, 2013 #4
    Oh. So this isnt squeeze action... I just find the limit of ln as sinx approaches zero. Hmm... I would need to graph that. But that should be simple enough. Tjanks guys
     
  6. Jan 27, 2013 #5

    Mark44

    Staff: Mentor

    Good idea. Remember that x is approaching π from the left, which says something about how sin(x) is approaching zero, and finally, what ln(sin(x)) is doing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the limit of ln (sinx) as x approaches pi
  1. Limit of ln(sinx) (Replies: 2)

Loading...