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anonymous

  • 5 years ago

Calculate the value of delta that corresponds to epsilon given limx->6 (2x+1) = 13, according to the definition of limits. Why delta = epsilon/2 is the only suitable solution? I have two cases: 1. If delta > epsilon/2 that's not going to work because our original |f(x)-L| will be greater than epsilon which is not possible: |f(x)-L| > epsilon 2. If delta < epsilon / 2 then |f(x)-L| < epsilon - a (where a < epsilon) which is ok since |f(x)-L| < epsilon - a < epsilon. So the first case is wrong. But what's about the second case? What if delta < epsilon/2? Will it still be wrong? Why?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. amistre64
    • 5 years ago
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    since this is a line; your delta is simple. :)

  3. amistre64
    • 5 years ago
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    lim{x->6} (2x+1) = 13 f(d) = 2d+1 = 13+e 2d = 12+e d = (12+e)/2 i think

  4. amistre64
    • 5 years ago
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    maybe.... es and ds tend to confuse me lol

  5. amistre64
    • 5 years ago
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    or should that be f(x+d)??

  6. amistre64
    • 5 years ago
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    d < e/2 is good

  7. amistre64
    • 5 years ago
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    i can see it, but trying to describe it eludes me....

  8. anonymous
    • 5 years ago
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    From formal definition of limit: \[\lim_{x \rightarrow a}f(x) = L\] \[|f(x) - L| < \epsilon\] So we have: \[|2x+1-13| < \epsilon\] \[|x-6| < \epsilon/2\] And also from formal definition of limit: \[0 < |x - a| < \delta\] Which gives us: \[0 < |x - 6| < \delta\] So delta can be anywhere between 0 and epsilon/2

  9. anonymous
    • 5 years ago
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    I would say that u can go to Paul's Online Math Notes for additional help it's a pretty good math website yall

  10. myininaya
    • 5 years ago
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    |2x+1-13|=|2x-12|=2*|x-6|<e solvinf this for |x-6| we can what we need to make d so |x-6|<e/2 so d=e/2

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