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FA07Ma1aFinalSols

FA07Ma1aFinalSols - Ma1a Final Solutions Fall 2007 Problem...

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Ma1a Final Solutions Fall 2007 Problem 1. Let f ( x ) be a real-valued function defined on R . Assume that f ( x ) is differentiable at x = 1, that is, f (1) exists. Prove that f ( x ) is continuous at x = 1. Solution 1. Analytic Solution Let ε > 0. We want to show that δ > 0 such that | x - 1 | < δ ⇒ | f ( x ) - f (1) | < ε . Set c = f (1). Then δ > 0 such that for 0 < | x - 1 | < δ , | f ( x ) - f (1) x - 1 - c | < 1 2 ε. Now c is a fixed number, so we can choose δ > 0 such that δ | c | < ε/ 2. Now let δ = min { 1 , δ , δ } , then if | x - 1 | < δ , | f ( x ) - f (1) - c ( x - 1) | < | x - 1 | 1 2 ε < 1 2 εδ 1 2 ε. Triangle inequality gives | f ( x ) - f (1) | ≤ | f ( x ) - f (1) - c ( x - 1) | + | c ( x - 1) | < 1 2 ε + | c ( x - 1) | ≤ 1 2 ε + δ | c | ≤ 1 2 ε + δ | c | < 1 2 ε + 1 2 ε = ε. Alternate Limit Solution To show that f is continuous at 1, it suffices to show that lim h 0 f (1 + h ) = f (1) . Note that for all h = 0, we may write this term as f (1 + h ) = f (1) + h · f (1 + h ) - f (1) h . Since f is differentiable at 1, the difference quotient will have a (finite) limit: f (1) = lim h 0 f (1 + h ) - f (1) h . Thus, taking a limit of the entire expression, we obtain lim h 0 f (1 + h ) = f (1) + 0 · f (1) = f (1) . Thus, f is continuous at 1. Problem 2. Is the series n =1 ( n 4 + 5 - n 2 ) absolutely convergent, conditionally but not absolutely convergent, or divergent? Explain the tests you use and their requirements. Solution 2. We have ( n 4 + 5 - n 2 ) = ( n 4 + 5 - n 2 )( n 4 + 5 + n 2 ) ( n 4 + 5 + n 2 ) = 5 n 4 + 5 + n 2 ( * ) 1
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2 Now, n 4 + 5 > n 2 n 4 + 5 + n 2 > 2 n 2 5 n 4 + 5 + n 2 < 5 2 n 2 Since 0 n 4 - 5 + n 2 5 2 n 2
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