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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
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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dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0this 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0}f(x)L<epsilon ? i have that written down

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0thanks...so thats whats given f(x) L = (x^2 5) +1 epsilon = .01

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep i need to find delta right?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ok i think its x 2 < delta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you get that, i think the equations copied wierd

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0i guessed...what is the correct equation with delta?? .01 < x^24 < .01 3.99<x^2 < 4.01 sqrt(3.99) < x < sqrt(4.01)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0delta < x2 < delta 2delta < x < 2+delta set them equal sqrt(3.99) = 2delta > delta = 2sqrt(3.99) = .002501 sqrt(4.01) = 2+delta > delta = sqrt(4.01)  2 = .002498
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