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Limit of ln(sinx)

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]lim_{x \to 0+} ln(sin(x))[/itex]

    [itex]lim_{x \to \infty} [ln(1+x^2)-ln(1+x)][/itex]

    2. Relevant equations

    3. The attempt at a solution
    I'm really not sure how to take this limit at all?

    I know (from using a table) that it tends towards -infinity, but I am not sure how to go about taking it manually?

    I am thinking that because as x approaches 0, sin(x) approaches 0, you can treat sin(x) like x. Then as x approaches 0, ln(x) approaches -infinity.

    For the second one, if I separate it into two limits, they both go towards infinity, then it's infinity - infinity?
    Last edited: Sep 1, 2011
  2. jcsd
  3. Sep 1, 2011 #2


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    Intuitively, that is correct. You can make it rigorous with a δ, ε argument. If you can use the fact that ln(x) → -∞ as x → 0+ then you know that given any N > 0 there is a δ > 0 such that ln(x) < -N if 0 < x < δ. Now put that together with a similar type of argument about sin(x), knowing that sin(x) → 0 as x → 0.
  4. Sep 1, 2011 #3


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    For the first one, I think the most rigorous approach (besides epsilon-delta) is the squeeze theorem. Use

    [tex]\sin(x)\leq x[/tex]

    Take logs and take the limit.

    For the second one, first make one logarithm using

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