Method 2. Since cosine has period 27, we may add -2πn to the argument of the cosine without changing its value; therefore, lim cos (27n²2 + 3n V 818 A =lim cos (27n² + 3n – 2πη = lim_cos (2π (√n² + 3n − n)) - 84x nx = lim cos (27 (√n²+3n-n)) = lim cos 818 n→∞ = cos 3π = -1. 3 (2= 2 + 0(1))

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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Can you please explain how the Taylor formula was applied I don’t fully understand how we got 3/2
7
Method 2.
Since cosine has period 27, we may add -2πn to the argument of the cosine without changing
its value; therefore,
lim cos(27√n² + 3n
84x
lim cos (27n²+ 3n
V
n4x
lim cos (2π (√√n²+3n − n)) =
n-x
n-x
– 2πη
πn) = lim cos (27 (√² + 3n_n))
cos (2π 32 +0(1))
lim_cos
n→∞
= cos 3π = –1.
DII
F12
Transcribed Image Text:7 Method 2. Since cosine has period 27, we may add -2πn to the argument of the cosine without changing its value; therefore, lim cos(27√n² + 3n 84x lim cos (27n²+ 3n V n4x lim cos (2π (√√n²+3n − n)) = n-x n-x – 2πη πn) = lim cos (27 (√² + 3n_n)) cos (2π 32 +0(1)) lim_cos n→∞ = cos 3π = –1. DII F12
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