# Statement(s) > [!NOTE] Theorem (Uniqueness of Limit of Convergent Real Sequence) > A [[Convergence]] can have at most one [[Convergence|limit]]. > [!NOTE] Theorem (Uniqueness of Limit of Convergent Sequence in $\mathbb{R}^n$) > Let $(x_{j})$ be [[Convergence|convergent sequence]] in $\mathbb{R}^n$ such that $x_{j} \longrightarrow_{j \to \infty} x$ and $x_{j} \longrightarrow_{j \to \infty} \bar{x}$ then $x = \bar{x}$. # Proof(s) ###### Proof Suppose $a_{n} \to l_{1}$ and $a_{n} \to l_{2}$ as $n \to \infty$. STS $|l_{1}-l_{2}|<\epsilon$ for all $\epsilon>0$ since [[There are no infinitesimal real numbers]]. Pick $\epsilon>0$. Then $\exists N_{1}, N_{2} \in \mathbb{N}$ such that $|a_{n}-l_{1}|<\epsilon$ and $|a_{n}-l_{2}|<\epsilon$ for all $n\geq N_{1}$ and $n \geq N_{2}$. Now for every $n \geq \max(N_{1}, N_{2})$, using [[Absolute Value Satisfies Triangle Inequality]], we have $|l_{1}-l_{2}|\leq |l_{1}-a_{n}|+|a_{n}-l_{2}| < 2\epsilon$ Hence $|l_{1}-l_{2}|=0 \implies l_{1}=l_{2} \quad \square$. ###### Proof of Uniqueness of Limit of Convergent Sequence in $\mathbb{R}^n$ \[MA270\]: BWOC, suppose $x \neq \bar{x}$, then let $\varepsilon:= \frac{1}{2} \lVert x - \bar{x} \rVert>0$. Since $x_{j}$ converges to $x$, there exists $N_{1}\in \mathbb{N}$ such that $j \geq N_{1} \implies \lVert x_{j} -x \rVert <\varepsilon .$Similarly, there exists $N_{2}\in \mathbb{N}$ such that $j\geq N_{2} \implies \lVert x_{j} - \bar{x} \rVert <\varepsilon. $Then applying [[Euclidean spaces are normed spaces|triangle inequality]], $2\varepsilon = \lVert x-\bar{x} \rVert \leq \lVert x - x_{j} \rVert + \lVert x_{j} - \bar{x} \rVert < 2\varepsilon. $This is of course a contradiction and therefore $x=\bar{x}$.