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Math Help - Definition of a Limit question.

  1. #1
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    Definition of a Limit question.

    When proving that

    \lim_{x\rightarrow p}(ax+b)=ap+b,a>0

    using the definition of the limit (or the -separated definition), find the largest possible value of when a=2 and =001
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  2. #2
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    \forall \epsilon > 0, \ \exists \delta > 0 such that whenever 0 < |x - p| < \delta it follows that |(ax + b) - (ap + b)| < \epsilon.

    Looking at the latter expression and noting that a > 0:
    |(ax + b) - (ap + b)| = |ax - ap| = |a(x-p)| = a|x - p| < \epsilon

    So what \delta should we choose?
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  3. #3
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    hmm, im lost
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    What don't you understand?

    Here's the formal definition of a limit:
    \forall \epsilon > 0, \ \exists \delta > 0 such that whenever 0 < |x - p| < \delta it follows that |f(x) - L| < \epsilon

    Here: f(x) = ax + b and L = ap + b.

    With these epsilon-delta proofs, we play around with the expression |f(x)-L| < \epsilon to get |x-p| < m for some m.

    Why do we do this? Because we know that |x - p| < \delta. So if we let \delta = a, then through all the algebra we did, it must necessarily follow that |f(x) - L| < \epsilon.

    By definition, this proves that \lim_{x \to p} f(x) = L
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  5. #5
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    But how would go about actually solving for \delta, ie the largest value when a=2 and \epsilon=0.01

    I guess I don't see where the answer lies in your explanation
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  6. #6
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    What I gave you was the way to get the solution to \delta in terms of both \epsilon and a. Just plug it in and it should work.
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