q4s - ² 1 4 k 1 = 1 1 4 k 1 Hence s n k → 1 where n k =...

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Math 310 - Quiz 4 solutions Wednesday, 4 Nov 2009 (20 pts) No calculators, notes or text allowed. Each problem is worth 10 points. 1. Write down the first five terms of the folowing sequence ( s n ) n . Find, if possible, a subsequence ( s n k ) k which converges to 1 (what is n k ?). s n := sin ± 2 ² + 1 n for n N . The first five terms of the sequence are as follows: s 1 = sin ± π 2 ² + 1 = 1 + 1 = 2 , s 2 = sin( π ) + 1 2 = 0 + 1 2 = 1 2 , s 3 = sin ± 3 π 2 ² + 1 3 = - 1 + 1 3 = - 2 3 , s 4 = sin(2 π ) + 1 4 = 0 + 1 4 = 1 4 , s 5 = sin ± 5 π 2 ² + 1 5 = 1 + 1 5 = 6 5 . For k N , set n k = 4 k + 1, then s n k = s 4 k +1 = sin ³ (4 k + 1) π 2 ´ + 1 4 k + 1 = sin ± π 2 + 2 ² + 1 4 k + 1 = sin ± π 2
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Unformatted text preview: ² + 1 4 k + 1 = 1 + 1 4 k + 1 . Hence, s n k → 1 where n k = 4 k + 1. ± 2. Let x ∈ R . Prove that the real-valued function f , defined by f ( x ) := 2 x + 3 for x ∈ R , is continuous at x by verifying the ε-δ property. Let ε > 0 and set δ := ε/ 2. Let x ∈ R and suppose that | x-x | < δ . Then we have | f ( x )-f ( x | = | 2 x + 3-2 x-3 | = 2 | x-x | < 2 δ = ε. Hence, f is continuous at x . ±...
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