# HW#11 - SOLUTIONS TO HOMEWORK 12 Problem A Discussion Since...

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Unformatted text preview: SOLUTIONS TO HOMEWORK # 12 Problem A . Discussion . Since | sin y | ≤ 1 and | sin y | = sin | y | ≤ | y | for each real y (the latter inequality was proved in the class), we have: | cos x- cos a | = 2 | sin x- a 2 || sin x + a 2 | ≤ 2 | x- a | 2 = | x- a | . Proof . Fix an arbitrary ε > 0. Let us put δ = ε . The above formula shows that if | x- a | < δ then | cos a- cos b | < ε . Q.E.D. 20.15 For x sufficiently large, ( x- 2) 3 > x 2 . Then 1 ( x- 2) 3 < 1 x 2 . Thus 0 ≤ fl fl fl fl 1 ( x- 2) 3 fl fl fl fl ≤ fl fl fl fl 1 x 2 fl fl fl fl , and the “policemen theorem” obviously gives lim x →∞ 1 ( x- 2) 3 = 0. Now fix m > 0 and k + δ- q 1 m . Then ( x- 2) < δ gives fl fl fl fl 1 ( x- 2) 3 fl fl fl fl ≥ 1 δ 3 = m . This by definition lim x → 2 + f ( x ) = ∞ . 20.16 (a) lim x → a + f 1 ( x ) = L 1 ⇒ ∀ ε > ∃ δ 1 > 0 s.t. x- a < δ 1 ⇒ fl fl fl fl f 1 ( x )- L 1 fl fl fl fl < ε/ 2....
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HW#11 - SOLUTIONS TO HOMEWORK 12 Problem A Discussion Since...

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