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Math Help - Limits using definition

  1. #1
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    Limits using definition

    Can you tell me how to find a limit using the definition only?
    For example
    Limit of 1/(5x-3) when x tends to 2.

    I do know how to prove the above limit is 1/7(if the value 1/7 is given) using the definition. But i do not know how to arrive at the value using the definition of limits

    Thanks
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  2. #2
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    Quote Originally Posted by Dili View Post
    Can you tell me how to find a limit using the definition only?
    For example
    Limit of 1/(5x-3) when x tends to 2.

    I do know how to prove the above limit is 1/7(if the value 1/7 is given) using the definition. But i do not know how to arrive at the value using the definition of limits

    Thanks
    Is this what you are looking for?

    \lim_{x \to 2^-}\frac{1}{5x - 3} = \frac{1}{7}
    and
    \lim_{x \to 2^+}\frac{1}{5x - 3} = \frac{1}{7}

    Since the limit exists on both sides and is the same we may unambigously state that
    \lim_{x \to 2}\frac{1}{5x - 3} = \frac{1}{7}

    If this isn't what you are looking for, I'm not sure what to say. The point x = 2 is in the domain of the expression, and as such we can usually say that the limit is just the value of the expression with x = 2. (Because of the argument above.)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Oh! Unless you mean you need an \epsilon - \delta proof? (I'd have to look that one up myself to make sure I've got the definitions right.)

    -Dan
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  4. #4
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    This is just an exercise in algebra.
    Note that: \left| {\frac{1}{7} - \frac{1}{{5x - 3}}} \right| = \left| {\frac{{5x - 10}}{{7\left( {5x - 3} \right)}}} \right|.

    If \left| {x - 2} \right| < 1 then \left| {\frac{1}{{7\left( {5x - 3} \right)}}} \right| < \frac{1}{{14}}

    Here it is. If \varepsilon  > 0 then choose \delta  = \min \left\{ {\frac{\varepsilon }{2},1} \right\}, then if 0 < \left| {x - 2} \right| < \delta we have \left| {\frac{{5x - 10}}{{7\left( {5x - 3} \right)}}} \right| = \frac{{5\left| {x - 2} \right|}}{{\left| {7\left( {5x - 3} \right)} \right|}} \le \frac{5}{{14}}\delta  < \varepsilon .

    QED
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  5. #5
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    Yes, by definition i meant the epsilon delta proof.
    But I do know how to prove lim 1/(5x-3) is 1/7 when x tends to 2.(the way Plato proved)
    The problem is we were given the limit, just to find the limit of 1/(5x-3) when x tends to 2 using the definition. We were not given the value 1/7, we have to arrive at it
    If anyone can help i would be really grateful

    Thanks
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  6. #6
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    Quote Originally Posted by Dili View Post
    The problem is we were given the limit, just to find the limit of 1/(5x-3) when x tends to 2 using the definition. We were not given the value 1/7, we have to arrive at it.
    I think that you have misunderstood the question.
    There really is no applying the  \varepsilon \& \delta method to find the limit.
    We use the definition to prove that 1/7 is the limit once we have found it by usual means.
    Last edited by Plato; August 18th 2007 at 05:27 AM.
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