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Math Help - limits proof

  1. #1
    Senior Member BAdhi's Avatar
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    limits proof

    prove from the definition whether limit exists or not

    \lim_{x \to 0} cos(\frac{1}{x})

    any idea how to start?
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  2. #2
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    As \displaystyle x \to 0, \frac{1}{x} \to \infty.

    The cosine function oscillates between \displaystyle [-1, 1], so there is not a limiting value.
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    If for every \epsilon >0 there exists a \delta >0 such that \left|\cos{\left(\frac{1}{x}\right)}-L\right|< \epsilon whenever 0< |x-0|< \delta.

    If we let cosine be sufficiently large, then cosine is either -1 or 1.

    \displaystyle \left|1-L\right|< \epsilon\Rightarrow \left|\frac{1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{1-L}

    \displaystyle \left|-1-L\right|< \epsilon\Rightarrow \left|\frac{-1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{-1-L}

    What do we let delta equal now?
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    If for every \epsilon >0 there exists a \delta >0 such that \left|\cos{\left(\frac{1}{x}\right)}-L\right|< \epsilon whenever 0< |x-0|< \delta.

    If we let cosine be sufficiently large, then cosine is either -1 or 1.

    \displaystyle \left|1-L\right|< \epsilon\Rightarrow \left|\frac{1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{1-L}

    \displaystyle \left|-1-L\right|< \epsilon\Rightarrow \left|\frac{-1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{-1-L}

    What do we let delta equal now?
    I don't understand what you mean by the bolded line, but your argument can not work - you have that |x|<\frac{x\epsilon}{-1-L} , but if x is positive then this means that a positive quantity (LHS) is smaller than a negative quantity (RHS), which is absurd.


    This is an outline of the proof I have in mind:

    1) Suppose towards a contradiction that the limit exists. Then since cosine is bounded, the limit is finite, say L.

    2) Find two sequences \{x_n\}, \ \{y_n\} such that x_n \to 0, \ y_n \to 0 as n \to \infty, and such that \forall n: \ cos(\frac{1}{x_n}) = 1 and cos(\frac{1}{y_n}) = 0, and argue why this finishes the proof.

    Edit: Notice that 2) was wrong at first, fixed it now.
    Last edited by Defunkt; December 23rd 2010 at 02:09 AM. Reason: fix
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  5. #5
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    "Let cosine be sufficiently large"??

    Proof by contradiction. Suppose the limit does exist. Let \lim_{x\to \infty} cos(x)= L. Then, given any \epsilon> 0, there exist R such that if x> R, |cos(x)- L|< \epsilon. Certainly, we must have -1\le L\le 1.

    Case 1: L\ge 0. Let \epsilon= 1- L> 0. There exist an integer n such that 3\pi n> R. For x= 3\pi n, cos(x)= cos(3\pi n)= -1 so [tex]|cos(x)- L|= L+1> 1- L, a contradiction.

    Case 2: L< 0. Let \epsilon= 1+ L> 0. Then there exist an integer n such that 2\pi n> R. You finish.
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  6. #6
    Senior Member BAdhi's Avatar
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    can you explain how took -1 <= L <= 1
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  7. #7
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    Quote Originally Posted by BAdhi View Post
    can you explain how took -1 <= L <= 1
    First, realize that the limit does not exist.
    But if it did exist, call it L, because |\sin(x)|\le 1 then  \displaystyle  \lim_{x\to 0}|\sin(\frac{1}{x})|\le 1..
    Thus it must be the case that -1\le L \le 1.
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  8. #8
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    Quote Originally Posted by BAdhi View Post
    can you explain how took -1 <= L <= 1
    Because -1\le cos(x)\le 1 for all x and a simple property of limits is that if a\le f(x)\le b then a\le \lim_{x\to x_0} f(x)\le b.

    Also, I notice now that the problem was \lim_{x\to 0} cos(1/x) and I did \lim_{\theta \to\infty} cos(\theta). That's not really a problem- just take \theta= 1/x.
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