Proof of Uniqueness of Limits

What the Uniqueness Proof Must Show

The Uniqueness of Limits theorem was established earlier in this course: if a real sequence converges to \(L\) and also to \(M\), then \(L=M\). This tutorial examines the proof in detail. The key idea is to suppose the proposed limits are different and choose a tolerance small enough that the sequence cannot be close to both at once.

The epsilon–N definition says that \(a_n\to L\) when, for every \(\varepsilon>0\), there is an index \(N\in\mathbb{N}_0\) such that \(|a_n-L|<\varepsilon\) whenever \(n\geq N\). If \(a_n\to L\) and \(a_n\to M\), then for any chosen positive tolerance, both estimates eventually hold. The thresholds for the two estimates may differ, but choosing the larger threshold makes them hold simultaneously.

If \(L\neq M\), their distance \(|L-M|\) is positive. Choosing a tolerance smaller than half that distance would require the same term \(a_n\) to lie in two disjoint neighborhoods. The proof succeeds because the tolerance is chosen from the distance between the proposed limits, rather than being arbitrary.

Separated Neighborhoods

A neighborhood of a real number \(x\), with radius \(r>0\), is the open interval \((x-r,x+r)\). In terms of distance, a real number \(y\) lies in this neighborhood exactly when \(|y-x|<r\). The following elementary fact makes the geometric idea in the uniqueness proof precise.

Proof. Since \(L\neq M\), the number \(d=|L-M|\) is positive. Set \(r=d/3\). Suppose, for contradiction, that some real number \(x\) belongs to both neighborhoods. Then \(|x-L|<r\) and \(|x-M|<r\). By the Triangle Inequality,

But \(|L-M|=d\), and \(d\) is not less than \(2d/3\) when \(d>0\). This contradiction proves that no \(x\) belongs to both neighborhoods. Therefore the neighborhoods are disjoint. \(\square\)

The specific choice \(d/3\) is convenient, not unique. Any positive radius \(r\) satisfying \(2r\leq d\) also gives disjoint neighborhoods. The proof above uses a strict inequality, so it works without needing to consider whether endpoints of the intervals are included: the neighborhoods are open, and membership requires distance strictly less than \(r\).

The Uniqueness Proof, Step by Step

Suppose a sequence \((a_n)\) converges to both \(L\) and \(M\). We give the proof by contradiction, making the choice of tolerance explicit. Assume \(L\neq M\), and let \(d=|L-M|>0\). Use the tolerance \(\varepsilon=d/3\). Convergence to \(L\) gives an index \(N_1\) such that \(|a_n-L|<d/3\) for every \(n\geq N_1\). Convergence to \(M\) gives an index \(N_2\) such that \(|a_n-M|<d/3\) for every \(n\geq N_2\). Let \(N\) be the larger of \(N_1\) and \(N_2\). For each \(n\geq N\), both inequalities hold. The Triangle Inequality then gives

This is impossible because \(|L-M|=d>2d/3\). Thus the assumption \(L\neq M\) is false, and \(L=M\). The proof depends on four points: the distance between distinct proposed limits is positive; the tolerance is chosen using that distance; the thresholds are combined; and the Triangle Inequality produces the contradiction. \(\square\)

This is a detailed examination of the proof of the Uniqueness of Limits theorem stated earlier, not a new version of that theorem. The neighborhood lemma describes the same contradiction geometrically: convergence would eventually place each term in both neighborhoods, while the neighborhoods are disjoint.

A Related Result: Distinct Limits Force Eventual Ordering

The same choice of a tolerance based on the gap between two limits proves a useful comparison result. It strengthens the geometric picture: not only can distinct limits be separated by neighborhoods, but two sequences converging to them must eventually appear in the same order as their limits.

Proof. First suppose \(L<M\), and write \(d=M-L\), so \(d>0\). Set \(\varepsilon=d/3\). Since \(a_n\to L\), there is an \(N_1\) such that, for every \(n\geq N_1\),

Since \(b_n\to M\), there is an \(N_2\) such that, for every \(n\geq N_2\),

Choose \(N\) to be the larger of \(N_1\) and \(N_2\). For \(n\geq N\), both bounds hold. Because \(d=M-L\),

since the right-hand side minus the left-hand side is \(d/3>0\). Consequently, \(a_n<b_n\) for every \(n\geq N\). If instead \(L>M\), apply the result just proved with the roles of \((a_n,L)\) and \((b_n,M)\) interchanged. This gives \(b_n<a_n\) for every sufficiently large \(n\). \(\square\)

For uniqueness, take the two convergent sequences in this theorem to be the same sequence: set \(b_n=a_n\) for every \(n\). If their proposed limits were distinct, the theorem would force either \(a_n<a_n\) or \(a_n>a_n\) for every sufficiently large \(n\). Both conclusions are impossible. This is another way to see how separation rules out two distinct limits.

Using the Proof to Check a Proposed Limit

The uniqueness proof is useful when a limit has been found by one method and a different answer is proposed by another. It does not require identifying which calculation is wrong: if both claims concern the same sequence and the proposed values differ, they cannot both be correct. A practical way to expose the conflict is to choose a tolerance that is less than half the distance between the claims.

A common pitfall is to choose a tolerance without relating it to the distance between the proposed limits. If the tolerance is too large, the two neighborhoods may overlap, and no contradiction follows. The proof does not require a uniquely correct choice: for distance \(d>0\), any positive \(\varepsilon\) with \(2\varepsilon\leq d\) is enough. Choosing \(d/3\) ensures the needed strict separation with a simple calculation.

Another pitfall is to overlook the thresholds. Convergence to \(L\) may provide one threshold, while convergence to \(M\) provides another. The contradiction applies only after both estimates hold, so the proof must take an index at least as large as both. This is why combining thresholds is part of the logic, not a merely notational step.

Finally, uniqueness concerns limits, not equality of terms. Two different sequences can converge to the same number, and a single sequence can have many terms far from its limit. What convergence guarantees is that all sufficiently late terms lie within any prescribed positive distance of the limit. When two proposed limits are distinct, a sufficiently small distance requirement makes those two demands incompatible.