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Continuity of sin x and cos x a. Use the identity sin (a +

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 93AE Chapter 2.6

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 93AE

Continuity of ? sin ?x and cos ?x a. Use the identi?ty si?n (a ? + ?? )=sin? ?a cos ? ? + co?? sin ?h with the fact that to prove that thereby establishing that ? sin ?x is continuo?us for a ? ll? . ?? in?t: L? ? = ?x ?? ?a? tha? t? = ?a + h? and note that ? h ? ? 0 as ? ? a.) b. Use th? e id?entity co ? s ? +?? ) = co? s ?a ? s ?h ? sin? sin h? with the fact that to prove that .

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STEP_BY_STEP SOLUTION Step-1 A continuous function can be formally defined as a f unction f : x y ,where the preimage of every open set in y is open in x. More concretely, a function f(x) in a single variable x is said to be continuous at point x if, 0 1. If f(x0) is defined, so that x 0 is in the domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x0 3. lx x(x) = f( x ).0 0 Left continuous : lim f() = f(a) , then f(x) is called a left continuous at x=a. xa Right continuous : lim f(x) = f(a) , then f(x) is called a right continuous at x=a. xa+ If , lim f(x) = f(a) = lim f(x) , then f(x) is called a continuous function at x=a. xa xa + If , f(x) is not continuous at x =a means , it is discontinuous at x=a. Step-2 a). Given that ; sin(a+h) = sin(a)cos(h) +cos(a)sin(h)..............(1) and limsin(x) = 0………(2) x0 Now , we have to prove that limsin(x) = sin(a) , by using the conditions (1) and(2). xa Given hint is ; Let h = x a so that x = a + h and note that h 0 as x a. limsin(x) = limsin(x-a+a) , since add and subtract a to x. xa xa Let , take h = x-a , That is , if x = a+h , then as x approaches to a , the value of h approaches to zero.[ since , a=a+h a-a = h 0 = h]. That is , xasin(x) = lxain(x-a+a) = lim sin(h +a) h0 = lim [sin(h)cos(a)+cos(h)sin(a)] , from (1). h0 = cos(a) lim sin(h) +sin(a)lim cos(h),since limCf(x) = Climf(x), h0 h0 xa xa here C is constant function. = cos(a) sin(0) +sin(a) cos(0), take the limits. = cos(a) (0) + sin(a) (1) , since sin(0) =0,and cos(0)=1. = sin(a). Therefore , limsin(x) = sin(a). xa Hence , sin(x) is continuous for all values of x. Step_3 NOTE: sin()is...

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Chapter 2.6, Problem 93AE is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

The answer to “Continuity of ? sin ?x and cos ?x a. Use the identi?ty si?n (a ? + ?? )=sin? ?a cos ? ? + co?? sin ?h with the fact that to prove that thereby establishing that ? sin ?x is continuo?us for a ? ll? . ?? in?t: L? ? = ?x ?? ?a? tha? t? = ?a + h? and note that ? h ? ? 0 as ? ? a.) b. Use th? e id?entity co ? s ? +?? ) = co? s ?a ? s ?h ? sin? sin h? with the fact that to prove that .” is broken down into a number of easy to follow steps, and 102 words. This full solution covers the following key subjects: sin, cos, use, prove, fact. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Since the solution to 93AE from 2.6 chapter was answered, more than 328 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 93AE from chapter: 2.6 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM.

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