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Prove that lim x sin -= 0. х p =x sin 2 Let f(x) be a function defined on an open interval about c, except possibly at c itself. Then the limit of f(x) as x approaches c is the number L, if, for every number e >0 there exists a corresponding number 8> 0 such that for all x, 0 < |x-c| <6 implies that f(x) – L| < ɛ. This limit is written as follows. lim f(x) = L Comparing the equation lim f(x) = L to the statement lim x sin corresponds to f(x). Therefore, prove the limit statement by showing that for every number ɛ > 0 there exists a 0, note that zero corresponds to c and L, and x sin х х х>0 corresponding number 8 >0 such that for all x, 0< |x| < & implies that x sin

Prove that lim x sin -= 0. х p =x sin 2 Let f(x) be a function defined on an open interval about c, except possibly at c itself. Then the limit of f(x) as x approaches c is the number L, if, for every number e >0 there exists a corresponding number 8> 0 such that for all x, 0 < |x-c| <6 implies that f(x) – L| < ɛ. This limit is written as follows. lim f(x) = L Comparing the equation lim f(x) = L to the statement lim x sin corresponds to f(x). Therefore, prove the limit statement by showing that for every number ɛ > 0 there exists a 0, note that zero corresponds to c and L, and x sin х х х>0 corresponding number 8 >0 such that for all x, 0< |x| < & implies that x sin

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Prove that lim x sin -= 0. х p =x sin 2 Let f(x) be a function defined on an open interval about c, except possibly at c itself. Then the limit of f(x) as x approaches c is the number L, if, for every number e >0 there exists a corresponding number 8> 0 such that for all x, 0 < |x-c| <6 implies that f(x) – L| < ɛ. This limit is written as follows. lim f(x) = L Comparing the equation lim f(x) = L to the statement lim x sin corresponds to f(x). Therefore, prove the limit statement by showing that for every number ɛ > 0 there exists a 0, note that zero corresponds to c and L, and x sin х х х>0 corresponding number 8 >0 such that for all x, 0< |x| < & implies that x sin <E. in f(x) = x sin Consider the factor sin The range of sin х is х х There does exist a function g(x) such that |f(x)|< |g(x)|. Notice that f(x) can not be bounded since its range is is as found in a previous step, the function f(x) satisfies f(x) = |x sin Since |f(x)| = |x sin |X| • sin and the range of sin х х х |<|x| <e → |f(x) – 0| <e. Now let ɛ be any positive number. Then 0< |x - 0| <ɛ → |x| <ɛ →
Transcribed Image Text:Prove that lim x sin -= 0. х p =x sin 2 Let f(x) be a function defined on an open interval about c, except possibly at c itself. Then the limit of f(x) as x approaches c is the number L, if, for every number e >0 there exists a corresponding number 8> 0 such that for all x, 0 < |x-c| <6 implies that f(x) – L| < ɛ. This limit is written as follows. lim f(x) = L Comparing the equation lim f(x) = L to the statement lim x sin corresponds to f(x). Therefore, prove the limit statement by showing that for every number ɛ > 0 there exists a 0, note that zero corresponds to c and L, and x sin х х х>0 corresponding number 8 >0 such that for all x, 0< |x| < & implies that x sin <E. in f(x) = x sin Consider the factor sin The range of sin х is х х There does exist a function g(x) such that |f(x)|< |g(x)|. Notice that f(x) can not be bounded since its range is is as found in a previous step, the function f(x) satisfies f(x) = |x sin Since |f(x)| = |x sin |X| • sin and the range of sin х х х |<|x| <e → |f(x) – 0| <e. Now let ɛ be any positive number. Then 0< |x - 0| <ɛ → |x| <ɛ →
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