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Material Type: Exam; Professor: Sharpley; Class: ANALYSIS I; Subject: Mathematics; University: University of South Carolina - Columbia; Term: Spring 2004;
Typology: Exams
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Math 554/703I - Analysis I Test 3 – April 14, 2004
Name:
Directions: To receive credit, you must justify your statements un- less otherwise stated. Answers should be provided in complete sentences.
1 (20 pts) 2 (10 pts) 3 (10 pts) 4 (10 pts) 5 (15 pts) 6 (10 pts) 7 (15 pts) 8 (10 pts)
(a) A closed set of real numbers that is not connected. One Soln: { 0 , 1 } is closed. (b) A set of real numbers that is complete but not connected. One Soln: [0, 1] ∪ [2, 3], a subset of the real numbers is complete iff it is closed. (c) A compact set of real numbers that is not connected. One Soln: [0, 1] ∪ [2, 3] is closed and bounded but is not connected. (d) A real-valued continuous function that does not satisfy the Intermediate Value Theorem. One Soln: Let dom(f ) = { 0 } ∪ [1, 2] and f (x) = x on that domain. The intermediate value of 12 is not attained. (e) A real-valued function that is continuous at a point x 0 , but is not differentiable. One Soln: The function f (x) = |x| is continuous at x 0 = 0, but is not differentiable there; observe sequential limits of the difference quotient applied to the sequences {− (^1) n } and { (^1) n }.
Soln: See course lecture notes as well as daily lecture notes.
x, then f ′(x 0 ) =
x 0
Proof: We just need to show the limit
xlim→x 0
x −
x 0 x − x 0
exists and equals L = 2 √^1 x 0. But
x − x 0 = (
x −
x 0 )(
x +
x 0 )
and so
xlim→x 0
x −
x 0 x − x 0
= lim x→x 0
x +
x 0 which by the limit theorems for quotients and sums, and the fact that the square root function is continuous at x 0 > 0 (and so limx→x 0
x =
x 0 ) implies that
xlim→x 0
x −
x 0 x − x 0
x 0