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Math Help - Another Epsilon Delta proof?

  1. #1
    Member mfetch22's Avatar
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    Another Epsilon Delta proof?

    Heres another problem that I haven't been able to figure out, I need some assistance:

    Suppose there is a \delta > 0 such that f(x) = g(x) when:

    0 < |x-a| < \delta

    Prove that:

    \lim_{x \to a} f(x) = \lim_{x \to a} g(x)
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  2. #2
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    Sounds like a very good teacher!

    I once taught a course called "Quantitative Methods for Economics and Business Administration" using a text book assigned by the Business Administration department. On one page of this book we were given three laws of limits:
    1) If \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M then \lim_{x\to a} (f+g)(x)= L+ M
    2) If \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M then \lim_{x\to a} fg(x)= LM
    3) if \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M and M is not 0, then \lim_{x\to a} \frac{f(x)}{g(x)}= \frac{L}{M}

    On the very next page, they introduced the derivative as the limit of the difference quotient: f'(a)= \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}, completely ignoring the fact that the denominator necessarily goes to 0 so this limit could NEVER be done using only those laws!

    Yes, that property is very important, and fairly easy to prove.

    Let \lim_{x\to a} f(x)= L. Then, by definition of limit, given any \epsilon> 0 there exist \delta_1> 0 such that if 0<|x- a|< \delta_1 then |f(x)- L|< \epsilon. Note the "<" symbol at the left of |x-a|. It does not matter what happens at x= a because then |x-a|= 0! (I am using the subscript on \delta_1 to distinguish if from the \delta given in the problem.)

    Now, given any \epsilon> 0, we can use the smaller of \delta and \delta_1 to prove that \lim_{x\to a} g(x)= L. If 0< |x- a|< the smaller of \delta and \delta_1, then both |f(x)- L|< \epsilon and f(x)= g(x) are true. So we can replace f(x) by g(x) in that expression: |g(x)- L|< \epsilon and we are done!
    Last edited by HallsofIvy; July 29th 2010 at 08:06 AM.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    hint:

    choose delta=epsilon
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    I once taught a course called "Quantitative Methods for Economics and Business Administration" using a text book assigned by the Business Administration department. On one page of this book we were given three laws of limits:
    1) If \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M then \lim_{x\to a} (f+g)(x)= L+ M
    2) If \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M then \lim_{x\to a} fg(x)= LM
    3) if \lim_{x\to a} f(x)= L and \lim_{x\to a} g(x)= M and M is not 0, then \lim_{x\to a} \frac{f(x)}{g(x)}= \frac{L}{M}

    On the very next page, they introduced the derivative as the limit of the difference quotient: f'(a)= \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}, completely ignoring the fact that the denominator necessarily goes to 0 so this limit could NEVER
    What you expected from "Quantitative Methods for Economics and Business Administration"?
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  5. #5
    Member Mathelogician's Avatar
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    If the limit for f is equal to L, then:
    There exists a delta>0 such that: 0<|x-a|<delta => |f(x)-L|<epsilon. since f(x)=g(x) => |g(x)-L|<epsilon.
    So we have: if 0<|x-a|<delta => |g(x)-L|<epsilon. Therefore the limit of g as x approaches a is L. So the limits at a are equal.
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  6. #6
    Member Mathelogician's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    hint:

    choose delta=epsilon
    Why equal?
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