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Unformatted text preview: [Take formal attendance] Campbell,David Young DeFilippo,James W Durkee,Zachary Martin Durrenberger,Marcelle Denise Gandevia,Munish Munir Graceffa,Erin R Kelly,Maureen Catherine Medeiros,Daniel Raposo Morris,John Patrick Theriault,Matthew Robert Vaughan,Stephen Douglas Veiga,Katti Michelle [Announce office hours.] Section 1.3: The notion of limit (concluded) Example: Prove lim x 1 2 x = 2. Proof: We need to find a ( ) such that 0 < | x 1| < ( ) implies |2 x 2| < . Rewrite this last inequality as 2 | x 1| < we see a simple winning strategy for Eve: = /2. Example: Prove lim x 1 x 2 = 1. Proof: We need to find a ( ) such that 0 < | x 1| < ( ) implies | x 2 1| < . Rewrite this last inequality as | x 1|| x + 1| < . If we knew that | x + 1| were at most 2, this would be just like the preceding example, and Eve could win by choosing = /2. But there is a way we can be sure that | x + 1| is at most 3, namely, by making sure that ( ) is 1 or smaller, because...
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This note was uploaded on 02/13/2012 for the course MATH 141 taught by Professor Staff during the Fall '11 term at UMass Lowell.
- Fall '11