Why Uniqueness Matters
A sequence may have many terms, and its early terms can look very different from its later terms. But if the terms eventually get arbitrarily close to a real number \(L\), there is no other real number they can eventually get arbitrarily close to as well. This fact is called the uniqueness of the limit.
Uniqueness matters whenever two calculations appear to give different limits. A graph or table may suggest a value, and an algebraic argument may suggest another. If both arguments really establish limits for the same sequence, the values must agree. If they do not, at least one argument contains an error.
Limits and Neighborhoods
Recall the epsilon-N definition for a real sequence \((a_n)\). The sequence converges to \(L\) if, for every \(\varepsilon>0\), there is a natural number \(N\) such that every term with \(n\geq N\) satisfies \(|a_n-L|<\varepsilon\). The notation \(a_n\to L\) means that the sequence converges to \(L\).
The epsilon-N definition says that, after some index, every term lies in any neighborhood of \(L\) that we specify. The index can depend on the chosen radius: a smaller neighborhood may require going farther along the sequence.
The Uniqueness Theorem
The key idea is to choose neighborhoods around \(L\) and \(M\) that do not overlap. If \(L\) and \(M\) are distinct, their distance is positive. Neighborhoods with sufficiently small radius cannot contain the same real number. But if the sequence converges to both values, all sufficiently late terms would have to lie in both neighborhoods at once.
Proof Using Disjoint Neighborhoods
Suppose, for a contradiction, that a sequence converges to two distinct real numbers \(L\) and \(M\). Let \(\delta=|L-M|\). Since \(L\ne M\), we have \(\delta>0\). Consider the open neighborhoods of radius \(\delta/3\) centered at \(L\) and \(M\).
These neighborhoods are disjoint. In fact, if some real number \(x\) belonged to both, then \(|x-L|<\delta/3\) and \(|x-M|<\delta/3\). The triangle inequality would give
But \(|L-M|=\delta\), so this is a contradiction. Therefore, no real number can belong to both neighborhoods.
Now use the assumption that \(a_n\to L\). By the definition of convergence, for the positive radius \(\delta/3\), there is an index \(N_1\) such that \(n\geq N_1\) implies \(|a_n-L|<\delta/3\). Likewise, because \(a_n\to M\), there is an index \(N_2\) such that \(n\geq N_2\) implies \(|a_n-M|<\delta/3\). Take \(N\) to be the larger of \(N_1\) and \(N_2\). For every \(n\geq N\), both inequalities hold, so \(a_n\) lies in both neighborhoods. That contradicts their disjointness. The assumption \(L\ne M\) must therefore be false, and \(L=M\).
Let \(L\ne M\), and write their positive distance as \(\delta=|L-M|\).
Use neighborhoods of radius \(\delta/3\); the triangle inequality shows that they are disjoint.
Beyond the larger of the two required indices, every term must lie in both neighborhoods.
A term cannot lie in two disjoint sets, so the supposed distinct limits cannot both be limits.
The proof does not require the sequence to be increasing, decreasing, or defined by a particular formula. It uses only the definition of convergence and the fact that distinct real numbers have a positive distance between them.
Worked Examples
Worked Example: Confirm a Limit and Reject a Second Candidate
Let \(a_n=4+\frac{1}{n}\), for \(n\in\mathbb{N}\). Verify that \(4\) is the limit, then test the claim that \(5\) is also a limit.
Step 1: Prove convergence to \(4\). Let \(\varepsilon>0\). Choose a natural number \(N>1/\varepsilon\). For every \(n\geq N\),
Thus the epsilon-N definition gives \(a_n\to 4\). The check is the defining inequality itself: the distance from the term to \(4\) is \(1/n\), and it is less than \(\varepsilon\) for all \(n\geq N\).
Step 2: Test the claim \(a_n\to 5\). For every \(n\geq 2\),
With \(\varepsilon=1/2\), no matter how large an index \(N\) is proposed, there is an \(n\geq N\) with \(n\geq 2\). For that term, the required strict inequality \(|a_n-5|<1/2\) fails. So \(5\) is not a limit. This agrees with uniqueness: the sequence has limit \(4\), and \(5\ne4\).
Worked Example: Check Competing Limits for a Rational Sequence
Consider \(a_n=\frac{3n+2}{n+1}\). Show that \(3\) is a limit and that \(2\) is not.
Step 1: Rewrite the difference from \(3\).
For any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(n\geq N\), then
This verifies \(a_n\to3\) directly from the definition.
Step 2: Reject \(2\). For every \(n\geq1\),
The last inequality follows because \(2n\geq n+1\) when \(n\geq1\). Taking \(\varepsilon=1/2\), every term fails the required strict inequality \(|a_n-2|<\varepsilon\). Thus \(2\) is not a limit. The calculation checks both claims: the difference from \(3\) shrinks to zero, whereas the difference from \(2\) stays at least \(1/2\).
Worked Example: Disprove a Nearby Limit Claim with Separated Neighborhoods
Let \(a_n=7+\frac{(-1)^n}{n}\). Verify that \(7\) is a limit, then show that \(69/10\) cannot be another limit.
Step 1: Verify convergence to \(7\). For every \(n\), \(|(-1)^n|=1\), so \(|a_n-7|=1/n\). Given \(\varepsilon>0\), choose \(N>1/\varepsilon\). Then \(n\geq N\) gives
Therefore \(a_n\to7\).
Step 2: Separate the proposed limits. Their distance is \(\left|7-\frac{69}{10}\right|=\frac{1}{10}\), so neighborhoods of radius \(\frac{1}{30}\) around these two numbers are disjoint. For \(n\geq60\), the distance to the second candidate satisfies
The inequality \(|u+v|\geq |u|-|v|\) gives the first bound. Thus, from \(n=60\) onward, the terms do not lie in the neighborhood of radius \(1/30\) around \(69/10\). This rules out \(69/10\) as a limit. It also illustrates the uniqueness argument: a sequence converging to \(7\) must eventually enter every neighborhood of \(7\), not a disjoint neighborhood centered at a different number.
A Common Mistake: Confusing the Relevant Difference
When checking a candidate limit, the expression to examine is the distance from the terms to that candidate. A small expression in the formula for \(a_n\) does not automatically measure that distance.
For example, take \(a_n=\frac{n}{n+1}\). A mistaken argument might say, “The difference from \(0\) is \(\frac{1}{n+1}\), which tends to zero, so the limit is \(0\).” The subtraction was done incorrectly. The actual difference is
For every \(n\geq1\), \(\frac{n}{n+1}\geq\frac12\), since \(2n\geq n+1\). Therefore the epsilon-N condition for a limit of \(0\) fails when \(\varepsilon=1/2\). The expression \(\frac{1}{n+1}\) is instead the distance to \(1\): \(|a_n-1|=\frac{1}{n+1}\). An exact subtraction check prevents a false limit claim.
Why the Result Is Useful
Uniqueness lets us treat a sequence's limit as a single definite value. Once a limit has been established, any different proposed value is automatically ruled out. This is useful when checking algebra, comparing two methods, or interpreting a table: numerical values may suggest a pattern, but a valid convergence argument must establish the defining inequalities.
The theorem does not say that every sequence has a limit. It says that if a real sequence has a limit, it cannot have two different real limits. Existence and uniqueness are separate questions: proving that a candidate is the limit establishes existence, while the uniqueness theorem prevents a second candidate from competing with it.
Check Your Understanding
Answer each question using the definition of a sequence limit or the uniqueness proof.
- In the uniqueness proof, why is \(\delta=|L-M|\) positive when \(L\ne M\)?
- Why does choosing neighborhoods of radius \(\delta/3\) guarantee that they are disjoint?
- If \(a_n\to L\) and \(a_n\to M\), why can one index be chosen after which every term satisfies both neighborhood conditions?
- For \(a_n=5+\frac{1}{n}\), show that \(5\) is a limit and explain why \(6\) cannot be a limit.
- Does uniqueness prove that every real sequence converges? Explain the distinction between existence and uniqueness.
