1 pt) Consider limx → 2 x2−5 = −1 .
Which of the following δ values guarantee that |(x2−5)+1|

Question

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

dumbcow · Answer

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

anonymous · Answer

}f(x)-L|<epsilon ? i have that written down

dumbcow · Answer

thanks...so thats whats given
f(x) -L = (x^2 -5) +1
epsilon = .01

anonymous · Answer

yep i need to find delta right?

dumbcow · Answer

yeah
ok i think its  |x- 2| < delta

anonymous · Answer

how did you get that, i think the equations copied wierd

dumbcow · Answer

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)

anonymous · Answer

okay you were right

dumbcow · Answer

-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