Math Problem Statement

prove the following theorem : When it exists, the limit of a sequence is unique

Solution

Theorem: When it exists, the limit of a sequence is unique.

Proof:

Let {an}\{a_n\} be a sequence, and suppose that this sequence converges to two distinct limits. Specifically, let:

limnan=L1andlimnan=L2.\lim_{n \to \infty} a_n = L_1 \quad \text{and} \quad \lim_{n \to \infty} a_n = L_2.

We aim to show that L1=L2L_1 = L_2, thus proving that the limit, if it exists, is unique.

Step 1: Definition of Limit

By the definition of a limit of a sequence, we know:

  1. Since limnan=L1\lim_{n \to \infty} a_n = L_1, for every ϵ>0\epsilon > 0, there exists an integer N1N_1 such that for all nN1n \geq N_1, we have:

anL1<ϵ.|a_n - L_1| < \epsilon.

  1. Similarly, since limnan=L2\lim_{n \to \infty} a_n = L_2, for every ϵ>0\epsilon > 0, there exists an integer N2N_2 such that for all nN2n \geq N_2, we have:

anL2<ϵ.|a_n - L_2| < \epsilon.

Step 2: Choose an Arbitrary ϵ>0\epsilon > 0

Now, let's consider an arbitrary positive number ϵ>0\epsilon > 0. We need to show that L1=L2L_1 = L_2.

Step 3: Choose N=max(N1,N2)N = \max(N_1, N_2)

Since both N1N_1 and N2N_2 exist, we can choose N=max(N1,N2)N = \max(N_1, N_2), which means that for all nNn \geq N:

anL1<ϵandanL2<ϵ.|a_n - L_1| < \epsilon \quad \text{and} \quad |a_n - L_2| < \epsilon.

Step 4: Use the Triangle Inequality

Now, consider the absolute difference between L1L_1 and L2L_2:

L1L2=(L1an)+(anL2).|L_1 - L_2| = |(L_1 - a_n) + (a_n - L_2)|.

By the triangle inequality, we have:

L1L2L1an+anL2.|L_1 - L_2| \leq |L_1 - a_n| + |a_n - L_2|.

For nNn \geq N, we know that both L1an<ϵ|L_1 - a_n| < \epsilon and anL2<ϵ|a_n - L_2| < \epsilon. Therefore:

L1L2ϵ+ϵ=2ϵ.|L_1 - L_2| \leq \epsilon + \epsilon = 2\epsilon.

Step 5: Conclusion

Since ϵ\epsilon is arbitrary, we can make it as small as we like. This implies that:

L1L22ϵfor everyϵ>0.|L_1 - L_2| \leq 2\epsilon \quad \text{for every} \quad \epsilon > 0.

Thus, L1=L2L_1 = L_2, which shows that the limit of a sequence, when it exists, is unique.

Final Conclusion:

We have proven that the limit of a sequence, when it exists, must be unique.


Would you like any more details or clarifications? Here are some related questions you might find useful:

  1. What is the formal definition of the limit of a sequence?
  2. Can a sequence have more than one limit? Under what conditions?
  3. How do the concepts of convergence and boundedness relate in sequences?
  4. What happens if a sequence does not converge?
  5. What is the difference between a limit point and a limit of a sequence?

Tip: Always remember that in the context of sequences, the limit is only defined if the terms of the sequence get arbitrarily close to a specific value as nn \to \infty. If they oscillate or fail to approach any particular value, the limit doesn't exist.

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Math Problem Analysis

Mathematical Concepts

Sequence Convergence
Limits of Sequences
Proof by Contradiction
Triangle Inequality

Formulas

|a_n - L_1| < epsilon
|a_n - L_2| < epsilon
|L_1 - L_2| <= |L_1 - a_n| + |a_n - L_2|

Theorems

Uniqueness of Limit for Sequences

Suitable Grade Level

Grades 11-12 (Advanced)