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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.1.141

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if L₁ and L₂ are numbers such that aₙ → L₁ and aₙ → L₂, then L₁ = L₂.

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Recall the definition of the limit of a sequence: A sequence \( a_n \) converges to a limit \( L \) if for every \( \varepsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( |a_n - L| < \varepsilon \).
Assume that the sequence \( a_n \) converges to two limits \( L_1 \) and \( L_2 \). This means that for every \( \varepsilon > 0 \), there exist \( N_1 \) and \( N_2 \) such that for all \( n > N_1 \), \( |a_n - L_1| < \varepsilon \), and for all \( n > N_2 \), \( |a_n - L_2| < \varepsilon \).
Let \( N = \max(N_1, N_2) \). Then for all \( n > N \), both inequalities hold simultaneously: \( |a_n - L_1| < \varepsilon \) and \( |a_n - L_2| < \varepsilon \).
Use the triangle inequality to relate \( L_1 \) and \( L_2 \): \[ |L_1 - L_2| = |L_1 - a_n + a_n - L_2| \leq |L_1 - a_n| + |a_n - L_2|. \] Since both terms on the right are less than \( \varepsilon \), we have \( |L_1 - L_2| < 2\varepsilon \).
Because \( \varepsilon > 0 \) is arbitrary, the only way for \( |L_1 - L_2| < 2\varepsilon \) to hold for all \( \varepsilon \) is if \( |L_1 - L_2| = 0 \), which implies \( L_1 = L_2 \). This proves the uniqueness of the limit.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definition of the Limit of a Sequence

A sequence {aₙ} converges to a limit L if, for every positive number ε, there exists an index N such that for all n > N, the terms aₙ are within ε of L. This formalizes the idea that the terms get arbitrarily close to L as n grows large.
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Proof by Contradiction

Proof by contradiction assumes the opposite of what you want to prove and shows this assumption leads to a logical inconsistency. In this case, assuming two different limits exist for the same sequence helps demonstrate that such a situation is impossible.

Triangle Inequality in Limit Proofs

The triangle inequality states that the absolute value of a sum is less than or equal to the sum of absolute values. It is used in limit proofs to relate distances between sequence terms and different limit candidates, helping to establish contradictions when assuming multiple limits.
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