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Unformatted text preview: Ma1a Final Solutions Fall 2007 Problem 1. Let f ( x ) be a realvalued 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 00 > 0 such that δ 00  c  < ε/ 2. Now let δ = min { 1 , δ , δ 00 } , 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 00  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 → f (1 + h ) = f (1) . Note that for all h 6 = 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 → f (1 + h ) f (1) h . Thus, taking a limit of the entire expression, we obtain lim h → 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....
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This note was uploaded on 02/22/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.
 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra, The Land

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