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Cosine limits Let n be a positive integer. Use graphical

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

Solution for problem 74RE Chapter 4

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 74RE

Let n be a positive integer. Use graphical and/ or analytical methods to verify the following limits:

a. \(\lim _{x \rightarrow 0}\ \frac{1-\cos x^n}{x^{2n}}=\frac{1}{2}\)                                              b. \(\lim _{x \rightarrow 0}\ \frac{1-\cos^n x}{x^2}=\frac{n}{2}\)

Step-by-Step Solution:

Solution Step 1 In this problem we have to verify the given limits either by using graphical method or by using analytical method. Let us use analytical method to verify the given limits. 1cos x 1 a. lim x2n = 2 x0 To verify this limit we will be using the following trigonometric formula. 1. cos + sin = 1 2 2 2 2. cos 2 = cos sin n Now consider lim 1cos x x0 x2n By using 2., cos x can be written as n 2 xn 2 xn cos x = cos ( 2 sin (2) …. (i) And by using 1., 1 can be written as n n 1 = cos 2 ( ) + sin 2 ( ) …. (ii) 2 2 1cos xn Using (i) and ( ii), expand the numerator of lim x2n as follows x0 n cos2( ) +sin2 ( ) (cos2( ) sin2 ( ) ) lim 1co2nx = lim 2 2 2n 2 2 x0 x x0 x n n n n cos2 (2) +sin (2) cos2 (2) +sin2( 2 = lim x2n x0 2sin ( ) = lim 2 x0 x2n 2sin (2) = lim n 2 x0 (x ) Multiply and divide by 4 in the denominator. We get, 2sin ( ) = lim 2 4(x )2 x0 4 2sin (2) = lim xn 2 x0 4( 2 2 sin (2) = 4im xn2 x0 (2) x = lim( sinn(2) 2 2 2 x0 sin x We know that lim x = 1. Thus we get x0 = (1) 2 = 1 2 1cos xn 1 Thus lim 2n = x0 x 2 Hence verified. Step 3 n b. lim 1co2 x = n x0 x 2 n Consider lim 1co2 x x0 x Substituting x = 0, we get n lim 1co2 x = 1 cos 0= which is an indeterminate form. x0 x 0 0 So we can apply l’hopital’s rule. l'Hôpital's Rule: f(x) 0 f(x) ± Suppose that we have one of the following cases, lim g(x) = 0r lim g(x)= ± xa xa Where a can be any real number, infinity or negative infinity. f(x) f(x) In these cases we have lim g(x)= lim gx) xa xa 1cos x dx(1cos x) lim x2 = lim d(x ) x0 x0 dx d n n1 n1 dx (1 cos x) = 0 ncos x ( sin x) = n cos x . sin x d (x ) = 2x dx d n 1cos x dx(1cos x) n cos1x . sin x Then lim x = lim dxx ) = lim 2x x0 x0 x0 n n1 sin x = l2mcos x . x x0 By using lim(f(x). g(x)) = limf(x) limg(x), we get xa xa xa = limcos n1x . lim sin x 2x0 x0 x We know that lim sin x= 1. Thus we get x0 x n n1 = 2os 0 (1) n = 2 1cos x n Thereforelix0 x = 2 Hence verified.

Step 2 of 3

Chapter 4, Problem 74RE is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Cosine limits Let n be a positive integer. Use graphical