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Lim log(sin(x))/log(x)

  1. Nov 28, 2012 #1
    Hello, i am new here and this is my first post, so please bear with me for any mistakes or so... and sorry for my english that is not so good
    now my question is about the solution of this limit: lim log(sin(x))/log(x) x->0, without using de l'hopital... thank you in advance
     
  2. jcsd
  3. Nov 28, 2012 #2

    mfb

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    Do you have an idea for the limit?
    You can replace sin(x) by an upper bound, and solve an easier problem afterwards.
     
  4. Nov 28, 2012 #3

    Curious3141

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    First you have to consider if the limit is even defined. For a limit to exist, the left and right handed limits have to exist, and they have to be equal. In this case, what happens with the log functions when you try to take the left-sided limit (i.e. as x tends to 0 from very small negative values)?
     
  5. Nov 28, 2012 #4

    mfb

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    It is a right-sided limit, obviously.
     
  6. Nov 28, 2012 #5

    Curious3141

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    And where is that stated? "Obviously"?
     
  7. Nov 28, 2012 #6

    mfb

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    It is so obvious that it is not necessary to write it here.

    To consider a left-sided limit, you would need a function which is defined there. While it is possible to define the logarithm of negative numbers, I do not expect that catapax works with complex numbers.

    ##f: \mathbb{R}^+ \to \mathbb{R}##, ##f(x)=\frac{log(sin(x))}{log(x)}## has a well-defined limit for ##x\to 0##. This is identical to the right-sided limit.
     
  8. Nov 28, 2012 #7
    Besides, the limit does exist even if you let x to be complex…

    Anyway, I think you could show this by using the definition of limits using epsilons and deltas. Start by guessing the correct value for the limit (it should be pretty easy to guess) and then prove it.
     
  9. Nov 28, 2012 #8
    i tried to solve the limit in this way, but i think it is not correct: logx=y,
    so lim log(sin(e))^y/y => lim log(sin(e)).... but it is wrong, the results should be 1
     
  10. Nov 28, 2012 #9

    Curious3141

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    When the limit is simply written as x -> 0 (as opposed to x -> 0+), I don't think it can simply be assumed that the right-sided limit is the one being asked for.

    Hence, it is invalid to assume that the limit exists simply because the right-sided limit exists.

    Yes, the left-sided limit exists if we extend the range of log(x) to the complex numbers. But that is not the usual unqualified definition of the log function.
     
  11. Nov 28, 2012 #10

    Curious3141

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    Anyway, *assuming the question is asking for the right-sided limit* (and therefore the limit can be assumed to exist), the most elementary way to solve this is to subtract [itex]\frac{\log x}{\log x} = 1[/itex] from the expression and see what it reduces to using algebra and the rules of logs. You still need to use the result [itex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/itex]. If you are not allowed to assume this result, you can prove it easily with the Taylor series for sin(x). Remember to add one back to what you get at the end to find the original limit.
     
  12. Nov 28, 2012 #11
    ok, i succeed doing it, thank you!
     
  13. Nov 28, 2012 #12

    mfb

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    Careful: I did not say that!
    In the general case, the limit applys to the whole neighborhood where the function is defined. If that is on the positive real axis only (like in this problem), the limit is identical to the right-sided limit.

    If the function is defined for positive real values only, both limits are identical.
     
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