Unformatted text preview:  (1 /x )2  < ±, or equivalently 1 / 2δ < x < 1 / 2 + δ, x 6 = 1 / 2 implies 2± < 1 /x < 2 + ±. (b) Find a δ that veriﬁes the required condition if ± = 0 . 1 . To get 1 . 9 < 1 /x < 2 . 1, need 1 / (2 . 1) < x < 1 / (1 . 9), so δ can be any positive number less than or equal to the smaller of 1 / 21 / (2 . 1) = 1 / 42 and 1 / (1 . 9)1 / 2 = 1 / 38, that is, any < δ ≤ 1 / 42. For example, δ = . 02 would do. 1...
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 Fall '08
 WILKENING
 Calculus, lim, student ID number

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