quizsol07

# quizsol07 - ) that con-verges to x , we have lim f ( x n )...

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Math 104, Solution to Quiz 7 Instructor: Guoliang Wu July 18, 2009 1. Are the following series convergent or divergent? Please justify your answers. (a) (7 points ) X n =2 ( - 1) n ln n Solution: The sequence a n = 1 ln n is nonnegative, decreas- ing (since ln n is increasing) and lim n →∞ 1 ln n = 0 . By the alternating series test, the series ( - 1) n a n con- verges. (b) (8 points) X n =2 1 n (ln n ) 2 . Solution: Let f ( x ) = 1 x (ln x ) 2 . Then f ( x ) is continuous, non- negative and decreasing (since x (ln x ) 2 is increasing) on [2 , ) . By the integral test, 1 n (ln n ) 2 converges if and only if R 2 f ( x ) dx < . Since Z 2 f ( x ) dx = Z ln 2 1 u 2 du < , where we made a substitution u = ln x, du = 1 x dx . Therefore the series 1 n (ln n ) 2 converges. 1

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2. (a) (5 points) Please state one version of the deﬁnition: “ f is continuous at x 0 dom( f ) .” Solution: If for any sequence ( x n ) in dom( f
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Unformatted text preview: ) that con-verges to x , we have lim f ( x n ) = f ( x ) , then f is said to be continuous at x dom( f ) . (b) (10 points) Prove that the function f below is continuous at x = 0 . f ( x ) = x cos( 1 x 2 ) , if x 6 = 0 , if x = 0 . Proof. Suppose ( x n ) is any sequence in dom( f ) = R that converges to x = 0 . Then f ( x n ) = x n cos( 1 x 2 n ) , if x n 6 = 0 , if x n = 0 . In either case,-| x n | f ( x n ) | x n | , for all n , since | cos( 1 x 2 ) | 1 for all x . By squeeze theorem, lim n f ( x n ) = 0 = f (0) , since lim | x n | = lim(-| x n | ) = 0 . Therefore, f is continuous at x = 0 . 2...
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## quizsol07 - ) that con-verges to x , we have lim f ( x n )...

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