Uniqueness of Sequence Limits

Why a Limit Must Be Unambiguous

A sequence is a list of real numbers, and convergence describes where its terms eventually gather. It would be difficult to use limits reliably if the same sequence could settle near two different real numbers. The key idea is that distinct numbers can be surrounded by neighborhoods that do not overlap. Convergence to each number would then require the same sufficiently late term to lie in both neighborhoods, which is impossible.

This result is called the uniqueness of sequence limits. It does not tell us how to find a limit; it tells us that once a sequence has a limit, no other real number can also be its limit.

Convergence and Neighborhoods

A real sequence is a function \(a:\mathbb{N}\to\mathbb{R}\), with terms written \(a_n\). Its limit is defined by requiring all sufficiently late terms to be as close as desired to a fixed real number.

Definition: A sequence \((a_n)\) converges to \(L\in\mathbb{R}\) if, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\). In this case, we write \(a_n\to L\).

For a center \(L\in\mathbb{R}\) and a radius \(r>0\), the open neighborhood of radius \(r\) around \(L\) is the interval \((L-r,L+r)\). Saying \(|x-L|<r\) is exactly the same as saying that \(x\) lies in this interval. Thus, convergence to \(L\) means that for every positive radius, all sufficiently late terms lie in the corresponding neighborhood of \(L\).

To prove uniqueness, suppose two proposed limits \(L\) and \(M\) are distinct. Their distance \(d=|L-M|\) is positive. Choosing neighborhoods with radius \(d/3\) gives two intervals that cannot overlap: a point in both would have to be less than \(d/3\) from each center, leaving the centers less than \(2d/3\) apart. But their distance is \(d\). This separation is the core of the proof.

d=|L-M|>0, \qquad U_{d/3}(L)=(L-d/3,L+d/3), \qquad U_{d/3}(M)=(M-d/3,M+d/3).

The Uniqueness Theorem

Theorem: If a real sequence converges to \(L\) and also converges to \(M\), then \(L=M\).

The theorem concerns any real sequence; its terms need not follow a particular formula. The proof assumes both convergence statements and shows that distinct limits would force a contradiction.

Assume the proposed limits differ.
Suppose \(a_n\to L\), \(a_n\to M\), and \(L\ne M\). Define \(d=|L-M|\), so \(d>0\).

Choose disjoint neighborhoods.
Use the positive radius \(\varepsilon=d/3\) around both \(L\) and \(M\). The two open neighborhoods cannot share a point.

Apply both convergence conditions at once.
Convergence to \(L\) gives an index bound \(N_L\); convergence to \(M\) gives a bound \(N_M\). For every \(n\ge\max(N_L,N_M)\), the term \(a_n\) must lie in both neighborhoods.

Reach a contradiction.
No point lies in both neighborhoods. The assumption \(L\ne M\) is therefore impossible, so \(L=M\).

Here is the complete argument, including why the neighborhoods are disjoint and why the two index bounds can be used together.

\begin{aligned} &\text{Suppose }a_n\to L,\quad a_n\to M,\quad\text{and }L\ne M.\\ &d=|L-M|>0,\qquad \varepsilon=\frac d3>0.\\ &\text{Since }a_n\to L,\text{ there is }N_L\in\mathbb{N} \text{ such that }n\ge N_L\Rightarrow |a_n-L|<\frac d3.\\ &\text{Since }a_n\to M,\text{ there is }N_M\in\mathbb{N} \text{ such that }n\ge N_M\Rightarrow |a_n-M|<\frac d3.\\ &\text{Set }N=\max(N_L,N_M).\text{ Then for every }n\ge N,\\ &\qquad |L-M|\le |L-a_n|+|a_n-M|<\frac d3+\frac d3=\frac{2d}{3}.\\ &\text{Thus }d<\frac{2d}{3},\text{ which contradicts }d>0. \end{aligned}

The inequality on the final lines is the triangle inequality for absolute values. The contradiction follows because \(2d/3<d\) whenever \(d>0\). Therefore the assumption that \(L\) and \(M\) were distinct must be false, and the limits are equal.

The same proof can be understood directly through the neighborhoods. For every \(n\ge N\), the two convergence conditions require \(a_n\in U_{d/3}(L)\) and \(a_n\in U_{d/3}(M)\). If a real number \(x\) belonged to both, then

|L-M|\le |L-x|+|x-M|<\frac d3+\frac d3=\frac{2d}{3}<d=|L-M|,

an impossibility. So the neighborhoods are disjoint, yet convergence would force a late term into their intersection. This is precisely the contradiction used in the theorem.

Worked Examples

Worked Example: Rejecting a Second Limit for \(3+1/n\)

Consider \(a_n=3+1/n\), and suppose someone claims that it converges to both \(3\) and \(4\). First, verify convergence to \(3\). Given any \(\varepsilon>0\), choose \(N=\lfloor 1/\varepsilon\rfloor+1\). Then \(N>1/\varepsilon\), so for every \(n\ge N\), \(1/n\le1/N<\varepsilon\). Hence

|a_n-3|=\left|3+\frac1n-3\right|=\frac1n<\varepsilon.

Now test the claim of convergence to \(4\) using \(\varepsilon=1/3\). If that claim were true, there would be an \(N_4\) such that \(n\ge N_4\) implies \(|a_n-4|<1/3\). Convergence to \(3\) also gives an \(N_3\) for this same radius. Choose \(n\ge\max(N_3,N_4,4)\). The condition for \(3\) gives \(|a_n-3|=1/n\le1/4<1/3\), while direct calculation gives

|a_n-4|=\left|3+\frac1n-4\right| =1-\frac1n \ge 1-\frac14=\frac34>\frac13.

This contradicts the required condition for convergence to \(4\). The two neighborhoods have centers one unit apart and radius \(1/3\), so they are disjoint. The calculation confirms that no sufficiently late term can satisfy both conditions. Thus the sequence converges to \(3\), not to \(4\).

Worked Example: Rejecting a Second Limit for \(-1+2/n\)

Let \(a_n=-1+2/n\). To verify that \(a_n\to-1\), take any \(\varepsilon>0\) and choose \(N=\lfloor 2/\varepsilon\rfloor+1\). For \(n\ge N\), \(n>2/\varepsilon\), and therefore

|a_n-(-1)|=\left|-1+\frac2n+1\right|=\frac2n<\varepsilon.

Suppose instead that the sequence also converges to \(0\). Use radius \(\varepsilon=1/3\). Both convergence claims would then hold for every \(n\) at least as large as the maximum of their two index bounds. Choose such an \(n\) with \(n\ge7\). The distance from \(-1\) is

|a_n-(-1)|=\frac2n\le\frac27<\frac13.

But for the same index, since \(2/n\le2/7<1\),

|a_n-0|=\left|-1+\frac2n\right| =1-\frac2n \ge 1-\frac27=\frac57>\frac13.

So the term satisfies the required closeness to \(-1\) but cannot satisfy the required closeness to \(0\). That contradicts the claim of convergence to \(0\). The calculation verifies the contradiction, while the theorem confirms that the actual limit \(-1\) is unique.

Worked Example: Comparing Two Widely Separated Proposed Limits

Suppose an argument claims that some sequence \((a_n)\) converges both to \(4\) and to \(-2\). The distance between the proposed limits is

$$ d=|4-(-2)|=6. $$

Choose radius \(d/3=2\). The neighborhood around \(4\) is \((2,6)\), and the neighborhood around \(-2\) is \((-4,0)\). These intervals are disjoint: every number in the first is greater than \(2\), while every number in the second is less than \(0\).

If both convergence claims held, there would be index bounds \(N_4\) and \(N_{-2}\) such that all \(n\ge N_4\) satisfy \(|a_n-4|<2\) and all \(n\ge N_{-2}\) satisfy \(|a_n+2|<2\). Take \(n=\max(N_4,N_{-2})\). Then \(a_n\) would have to belong to both \((2,6)\) and \((-4,0)\). Since the intervals have no common point, this is impossible. The proposed pair of limits is therefore inconsistent. This check relies only on the two claims of convergence; it does not require knowing the sequence’s formula.

A Common Pitfall: Choosing a Radius That Is Too Large

To create a contradiction, the neighborhoods around distinct proposed limits must actually be disjoint. A positive radius by itself is not enough. For example, if the proposed limits are \(0\) and \(2\), choosing radius \(2\) gives the intervals \((-2,2)\) and \((0,4)\). These overlap on \((0,2)\), so it is not contradictory for a term to lie in both.

The correct choice depends on the distance \(d=|0-2|=2\). Taking radius \(d/3=2/3\) gives

\left(-\frac23,\frac23\right) \qquad\text{and}\qquad \left(\frac43,\frac83\right),

which do not overlap. More generally, any radius \(r\) with \(0<r<d/2\) works. The choice \(d/3\) is convenient because it is always positive and leaves a clear gap between the neighborhoods.

A second error is to obtain two index bounds but treat them as if they automatically refer to the same terms. If convergence to \(L\) gives \(N_L\) and convergence to \(M\) gives \(N_M\), use \(N=\max(N_L,N_M)\). Every \(n\ge N\) then satisfies both conditions. This common bound is essential to the contradiction.

Why Uniqueness Matters

Uniqueness makes limit notation meaningful. If a sequence has a limit, we can refer to the limit without ambiguity. The theorem also provides a useful check on calculations: if two methods appear to give different limits for one sequence, at least one method must contain an error.

Notice the scope of the result. It proves that there cannot be two different limits; it does not prove that every sequence converges. A sequence may fail to converge altogether. Uniqueness applies only once convergence to a real number has been established.

Takeaway: If \(a_n\to L\) and \(a_n\to M\), choose neighborhoods around \(L\) and \(M\) with radius smaller than half their distance. The neighborhoods are disjoint, but a common sufficiently late term would have to lie in both. Therefore \(L=M\).

Check Your Understanding

Use the definition of convergence and the neighborhood argument to answer these questions.

State the epsilon-\(N\) definition of \(a_n\to L\).
If \(L\ne M\), why is \(d=|L-M|\) positive, and what radius \(r\) guarantees disjoint neighborhoods?
In the uniqueness proof, why do we choose \(N=\max(N_L,N_M)\)?
For proposed limits \(1\) and \(5\), calculate their distance and give a radius that produces disjoint neighborhoods.
Why does choosing a radius larger than half the distance between the centers fail to guarantee disjoint neighborhoods?

What You'll Learn