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# sols23 - 54 23 November 29th Continuity and Topology 23.1...

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Unformatted text preview: 54 23. November 29th: Continuity and Topology. 23.1 . Verify formally that lim x → 1 x 2 = 1. Answer: We need to verify that ∀ > , ∃ δ > 0 : 0 < | x- 1 | < δ = ⇒ | x 2- 1 | < . Given , let δ = min( 3 , 1). If | x- 1 | < min 3 , 1 then in particular | x- 1 | < 1, so 0 < x < 2, and it follows that 1 < x + 1 < 3: so | x + 1 | < 3 . But then we have | x- 1 | < 3 and | x + 1 | < 3, and hence | x 2- 1 | = | x- 1 | · | x + 1 | < 3 · 3 = , as needed. 23.2 . Prove Lemma 23.1. Answer: We have to verify that if ( † ) ∀ > , ∃ δ > 0 : 0 < | x- c | < δ = ⇒ | f ( x )- L | < and ( ‡ ) ∀ > , ∃ δ > 0 : 0 < | x- c | < δ = ⇒ | f ( x )- L | < then L = L . Informally: Statement ( † ) says that f ( x ) is as close as we want to L for all x close enough to c ; statement ( ‡ ) says that f ( x ) is as close as we want to L for the same x . But if L and L are not the same number, f ( x ) cannot be arbitrarily close to both of them at the same time. Therefore necessarily L = L ....
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## This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.

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sols23 - 54 23 November 29th Continuity and Topology 23.1...

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