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anonymous

  • 4 years ago

1 pt) Consider limx → 2 x2−5 = −1 . Which of the following δ values guarantee that |(x2−5)+1| < .01 ? There may be more than one correct answer. Choose all correct answers. A. 0.003322 B. 0.002498 C. 0.002337 D. 0.002609 E. 0.002576

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  1. dumbcow
    • 4 years ago
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    this comes from the formal definition of the limit, i forget what it is though. could you write the general equations or i can look them up

  2. anonymous
    • 4 years ago
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    }f(x)-L|<epsilon ? i have that written down

  3. dumbcow
    • 4 years ago
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    thanks...so thats whats given f(x) -L = (x^2 -5) +1 epsilon = .01

  4. anonymous
    • 4 years ago
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    yep i need to find delta right?

  5. dumbcow
    • 4 years ago
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    yeah ok i think its |x- 2| < delta

  6. anonymous
    • 4 years ago
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    how did you get that, i think the equations copied wierd

  7. dumbcow
    • 4 years ago
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    i guessed...what is the correct equation with delta?? -.01 < x^2-4 < .01 3.99<x^2 < 4.01 sqrt(3.99) < x < sqrt(4.01)

  8. anonymous
    • 4 years ago
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    okay you were right

  9. dumbcow
    • 4 years ago
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    -delta < x-2 < delta 2-delta < x < 2+delta set them equal sqrt(3.99) = 2-delta --> delta = 2-sqrt(3.99) = .002501 sqrt(4.01) = 2+delta --> delta = sqrt(4.01) - 2 = .002498

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