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Unformatted text preview: 2, then 2  n is positive, so T = ( 1 + x 2 n 2 + Ö(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 + Ö(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 + Ö(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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 Spring '10
 AlbertoDelgado
 Combinatorics, Hypotenuse, triangle, sy T= sx, right triangle POR, Philippe Fondanaiche

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