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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4 - Problem 74re
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4 - Problem 74re

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# Cosine limits Let n be a positive integer. Use graphical ISBN: 9780321570567 2

## Solution for problem 74RE Chapter 4

Calculus: Early Transcendentals | 1st Edition

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Problem 74RE

Cosine limits? Let n be a positive integer. Use graphical and/or analytical methods to verify the following limits: 1?cos x 1 a. lim xn = 2 x?0 n b. lim 1?co2 x = n x?0 x 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.

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