3 Real Analysis II Section 17 and 18 David Joseph Stith b lim n 2 1 n 0 Suppose

# 3 real analysis ii section 17 and 18 david joseph

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Unformatted text preview: 3 Real Analysis II: Section 17 and 18 David Joseph Stith (b) lim( √ n 2 + 1 − n ) = 0 . Suppose ǫ > 0. We will show that n > 1 ǫ = ⇒ vextendsingle vextendsingle ( √ n 2 + 1 − n ) − vextendsingle vextendsingle < ǫ and hence lim( √ n 2 + 1 − n ) = 0. Suppose n > 1 ǫ . Then, n > 1 ǫ = ⇒ n 2 > 1 ǫ 2 = ⇒ n 2 + 1 > 1 ǫ 2 since n 2 + 1 > n 2 = ⇒ vextendsingle vextendsingle vextendsingle radicalbig n 2 + 1 vextendsingle vextendsingle vextendsingle > 1 ǫ = ⇒ vextendsingle vextendsingle vextendsingle radicalbig n 2 + 1 + n vextendsingle vextendsingle vextendsingle > 1 ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle 1 √ n 2 + 1 + n vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle √ n 2 + 1 − n ( √ n 2 + 1 + n )( √ n 2 + 1 − n ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle radicalbig n 2 + 1 − n vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle ( radicalbig n 2 + 1 − n ) − vextendsingle vextendsingle vextendsingle < ǫ Therefore lim( √ n 2 + 1 − n ) = 0. Q.E.D. (c) lim( √ n 2 + n − n ) = 1 2 Suppose ǫ > 0. We will show that n > 1 16 ǫ 2 = ⇒ vextendsingle vextendsingle ( √ n 2 + n − n ) − 1 2 vextendsingle vextendsingle < ǫ and hence lim( √ n 2 + n − n ) = 1 2 . Suppose n > 1 16 ǫ 2 . Then, 4 Real Analysis II: Section 17 and 18 David Joseph Stith n > 1 16 ǫ 2 = ⇒ n 2 + n > 1 16 ǫ 2 since n 2 + n > n = ⇒ radicalbig n 2 + n > 1 4 ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle radicalbig n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle > 1 4 ǫ since n + 1 2 > = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 √ n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < 4 ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle − 1 4 √ n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle n 2 + n − ( n 2 + n + 1 4 ) √ n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ( n 2 + n ) − ( n 2 + 1 2 ) 2 √ n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle [ √ n 2 + n − ( n + 1 2 )][ √ n 2 + n + ( n + 1 2 )] √ n 2 + n + ( n + 1 2 ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle ( radicalbig n 2 + n − n ) − 1 2 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ Therefore lim( √ n 2 + n − n ) = 1 2 . Q.E.D....
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• Fall '08
• Akhmedov,A
• Mathematical analysis, Tn, Dominated convergence theorem, David Joseph Stith

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