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Evaluate the limit. 12 lim х — 5 - x-5 x2 + 2x – 35 Give your answer in exact form. If the limit does not exist, enter DNE. lim 5-+ 2x - 35 1 12 x-5 х — x2 + 2x – 35 Select the statement that justifies your answer. 1 12 The limit exists because lim x-5 = L, where L is a real number. x2 + 2x – 35 12 The limit does not exist because lim = 00. x-5 x – 5 x2 + 2x – 35 12 The limit does not exist because lim = -00. x-5 x2 + 2x – 35 12 The limit does not exist because lim x-5 is undefined. х x2 + 2x – 35

Evaluate the limit. 12 lim х — 5 - x-5 x2 + 2x – 35 Give your answer in exact form. If the limit does not exist, enter DNE. lim 5-+ 2x - 35 1 12 x-5 х — x2 + 2x – 35 Select the statement that justifies your answer. 1 12 The limit exists because lim x-5 = L, where L is a real number. x2 + 2x – 35 12 The limit does not exist because lim = 00. x-5 x – 5 x2 + 2x – 35 12 The limit does not exist because lim = -00. x-5 x2 + 2x – 35 12 The limit does not exist because lim x-5 is undefined. х x2 + 2x – 35

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter7: Eigenvalues And Eigenvectors
Section7.4: Applications Of Eigenvalues And Eigenvectors
Problem 10E: Find the limit if it exists of Anx1 as n approaches infinity, where A=[02120], and x1=[aa]
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Evaluate the limit. 1 12 lim x→5 х — 5 x2 + 2x – 35 Give your answer in exact form. If the limit does not exist, enter DNE. 1 12 lim x-5 x2 + 2x 35 Select the statement that justifies your answer. 1 12 The limit exists because lim = L, where L is a real number. x→5 х — 5 x2 + 2x 35 1 12 The limit does not exist because lim = 0. x-5 x2 + 2x 35 1 12 The limit does not exist because lim x→5 \ x = -0. х2 + 2х — 35 1 12 35) is undefined. The limit does not exist because lim | x→5 х — 5 x2 + 2x
Transcribed Image Text:Evaluate the limit. 1 12 lim x→5 х — 5 x2 + 2x – 35 Give your answer in exact form. If the limit does not exist, enter DNE. 1 12 lim x-5 x2 + 2x 35 Select the statement that justifies your answer. 1 12 The limit exists because lim = L, where L is a real number. x→5 х — 5 x2 + 2x 35 1 12 The limit does not exist because lim = 0. x-5 x2 + 2x 35 1 12 The limit does not exist because lim x→5 \ x = -0. х2 + 2х — 35 1 12 35) is undefined. The limit does not exist because lim | x→5 х — 5 x2 + 2x
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