Explanation Given information: The given inequality implication is if 0 < | x − 3 | < δ then | 1 x − 1 3 | < 0.2 . Concept used: For the limit statement lim x → c f ( x ) = L , the delta-epsilon statement is For all ∈ > 0 , there exist δ > 0 such that whenever x ∈ ( c − δ , c ) ∪ ( c , c + δ ) guarantees f ( x ) ∈ ( L − ∈ , L + ∈ ) . Mathematically it can be written as for every x satisfying 0 < | x − c | < δ , every f ( x ) satisfies | f ( x ) − L | < ∈ . Calculation: Consider the given inequality. 0 < | x − 3 | < δ ......... (1) | 1 x − 1 3 | < 0.2 .......... (2) Now, solve the inequality (2) as shown: | 1 x − 1 3 | < 0

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 1.3, Problem 24S

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To find: the largest possible value of δ or smallest possible value of N that makes the given implication true. Verify the answer by graph.

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Chapter 1.3, Problem 23S

Chapter 1.3, Problem 25S

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Chapter 1 Solutions

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the largest possible value of δ or smallest possible value of N that makes the given implication true. Verify the answer by graph.

Solution Summary: The author explains that the largest possible value of delta is 0.66.

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To find: the largest possible value of δ or smallest possible value of N that makes the given implication true. Verify the answer by graph.

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Chapter 1 Solutions