The Uniqueness of Limits of a Sequence Theorem

# The Uniqueness of Limits of a Sequence Theorem

Recall that a given sequence $(a_n)$ is said to be convergent to some real number $L$ if $\lim_{n \to \infty} a_n = L$. We have not yet verified that $L$ is unique, that is - is it possible for a sequence to converge to two different real numbers $L$? The answer is no, and in fact if $(a_n)$ is convergent then $L$ is unique. The following theorem and proof will verify this assertion.

Theorem (Uniqueness of Limits of a Sequence): If $(a_n)$ is a convergent sequence to some real number $L$ then $\lim_{n \to \infty} a_n = L$ is unique, that is if $\lim_{n \to \infty} a_n = M$ then $L = M$. |

**Proof:**Let $(a_n)$ be a convergent sequence, and suppose that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} a_n = M$ where $L, M \in \mathbb{R}$ and $L \neq M$. Also, without loss of generality assume $L < M$.

- Now consider $\epsilon_1 = \frac{\mid L - M \mid}{2} > 0$.

- We note that $\lim_{n \to \infty} a_n = L$ implies that for $\epsilon_1$ there exists a natural number $N_1 \in \mathbb{N}$ such that $\forall n ≥ N_1$ then $x_n \in V_{\epsilon_1} (L)$.

- Similarly $\lim_{n \to \infty} a_n = M$ implies that for $\epsilon_1$ there exists a natural number $N_2 \in \mathbb{N}$ such that $\forall n ≥ N_2$ then $x_n \in V_{\epsilon_1} (M)$.

- Now choose $N = \mathrm{max} \{ N_1, N_2 \}$ such that both conditions above are satisfied. Then $\forall n ≥ N$ we have that $x_n \in V_{\epsilon_1} (L) \cap V_{\epsilon_1} (M)$. But we note that $V_{\epsilon_1}(L) \cap V_{\epsilon_1} (M) = \emptyset$ because $V_{\epsilon_1}(L) = \{ x_n \in \mathbb{R} : L - \frac{\mid L - m \mid}{2} < x_n < L + \frac{\mid L - M \mid}{2} \}$ and $V_{\epsilon_1}(M) = \{ x_n \in \mathbb{R} : M - \frac{\mid L - M \mid}{2} < x_n< M + \frac{\mid L - M \mid}{2} \}$ have no points of intersection as illustrated below:

- Therefore $L = M$, and so $L$ is unique. $\blacksquare$