Uniqueness of a Convergent Sequence's Limit

Suppose that $\{ a_n \}$ is a convergent sequence, that is as $n \to \infty$, $a_n \to L$. We will now prove that this limit $L$ is unique, but before that, we will first look at the following lemma necessary to prove this.

Lemma: If $x$ is a number such that $\forall \epsilon > 0$, $\mid x \mid < \epsilon$, then $x = 0$.

• Proof of Lemma: We note that $0 ≤ \mid x \mid$ and $x < \epsilon$. Therefore $0 ≤ \mid x \mid < \epsilon$. But for ANY $\epsilon > 0$, $\mid x \mid < \epsilon$. The only value of $x$ for which this is always true is $x = 0$, that is $0 ≤ \mid 0 \mid < \epsilon$. $\blacksquare$

We will now look at proving that a convergent sequence's limit is unique.

Theorem: If $\{ a_n \}$ is a sequence that converges at both $L$ and $M$, that is $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} a_n = M$, then $L = M$ or rather, the limit of this sequence is unique.

• Proof of Theorem: Assume that as $n \to \infty$ then $a_n \to L$ AND $a_n \to M$.

• Let $\epsilon > 0$ be given and choose $N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - L \mid < \frac{\epsilon}{2}$. Furthermore, choose $N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid a_n - M \mid < \frac{\epsilon}{2}$.

• Now choose $N = \mathrm{max} ( N_1, N_2 )$. Note that if we choose the maximum of $N_1$ and $N_2$, then it will be true that regardless $\mid a_n - L \mid < \frac{\epsilon}{2}$ and $\mid a_n - M \mid < \frac{\epsilon}{2}$. Now consider the following inequality:

(1)

\begin{align} \quad \mid L - M \mid = \mid L - a_n + a_n - M \mid = \mid (L - a_n) + (a_n - M) \mid \end{align}

• By the triangle inequality ($\mid A + B \ \mid ≤ \mid A \mid + \mid B \mid$), we get that:

(2)

\begin{align} \quad \mid L - M \mid = \mid (L - a_n) + (a_n - M) \mid ≤ \mid L - a_n \mid + \mid a_n - M \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• Therefore $\mid L - M \mid < \epsilon$. Now we note that by Lemma 1, that $L - M$ is a NUMBER, and $\forall \epsilon > 0$, $\mid L - M \mid < \epsilon$. This implies that $L - M = 0$ or rather $L = M$.

• So if as $n \to \infty$, $a_n \to L$ and $a_n \to M$, then $L = M$. We have showed that the limit of a convergent sequence is unique. $\blacksquare$