This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 OLSAVSKYGREGORM

Click to edit the document details