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Math Help - Epsilon-Delta

  1. #1
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    Epsilon-Delta

    Express the following as an Epsilon-Delta proof (to show that it is continuous):

    \lim_{x\rightarrow - \infty}f(x) = L

    Can someone give me ideas for this one?
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  2. #2
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    Quote Originally Posted by Jacobpm64 View Post
    Express the following as an Epsilon-Delta proof (to show that it is continuous):

    \lim_{x\rightarrow - \infty}f(x) = L

    Can someone give me ideas for this one?
    Continous has nothing to do with that.

    But it means.
    For any \epsilon >0
    There exists a N.
    If,
    x<N is in the domain
    Then,
    |f(x)-L|<\epsilon
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  3. #3
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    all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

    So, usually I'd write
    |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon

    Then I'd try to get that in the form |x - c| < \delta

    How do I do that here?
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  4. #4
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    Quote Originally Posted by Jacobpm64 View Post
    all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

    So, usually I'd write
    |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon

    Then I'd try to get that in the form |x - c| < \delta

    How do I do that here?
    There is no delta here. It is a different type of limit.
    Maybe you mean this.
    \forall \epsilon >0, \exists N \, \, \, \, \, x<N\to |f(x)-L|<\epsilon
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  5. #5
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    Oh, all right.. No delta because x is extending without bound towards negative infinity, but there is an epsilon because the function approaches a finite value of L, correct?

    I understand that.

    But, I still don't know how I'd go about constructing a proof.
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  6. #6
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    Quote Originally Posted by Jacobpm64 View Post

    But, I still don't know how I'd go about constructing a proof.
    Say you want to show,
    \lim_{x\to -\infty} \frac{1}{x}=0

    You need to show that for any \epsilon>0
    We can choose an N
    Such that if,
    x<N is in the domain of the function.
    Implies,
    \left|\frac{1}{x}\right|<\epsilon
    Since both side are positive we can take the reciprocal of both sides, but then we have to flip the inequality, obtaining an equivalent statement,
    |x|<\frac{1}{\epsilon}
    That means, x<\frac{1}{\epsilon}.
    We also note that,
    \frac{1}{\epsilon} is in the domain of f(x).

    So this tell us given any \epsilon we need to choose N=\frac{1}{\epsilon}
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  7. #7
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    wow, major headache.. I understand all your examples

    Thing is. The only thing that's killing me on mine is that there is no specific function... I'm just given f(x)

    In your example, you wrote |f(x) - L| < \epsilon

    as  x < \frac{1}{ \epsilon }

    That isn't too bad to do because f(x) is defined as a specific function,  \frac{1}{x}

    I don't know how I'd write  |f(x) - L| < \epsilon in the form x < something, when I don't have any specific function, so I only have an f(x), not an x.

    *sigh*
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