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Unformatted text preview: 2, then 2  n is positive, so T = ( 1 + x 2 n 2 + (1 + x 2 n 2 ) ) x 2  n with the first term going to 2 and the second term going to 0, as x 0. So T in this case. If n > 2, the term 1 + (1 + x 2 n 2 ) x 2 n 2 since the numerator approaches 2, as x goes to 0, while the denominator gets arbitrarily small. On the other hand x 2 n 2 x n 2 = x n 0 as x 0. So in this case T . Finally, if n = 2, then T = 1 + x 2 + (1 + x 2 ) 2. Finally we have the answer. T 0, if n < 2 2, if n = 2 if n > 2 This proves, I guess, that 2 is between 0 and . this page....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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