[proofplan] We prove uniqueness by contradiction. If the two proposed [limits](/page/Limit) $l$ and $k$ are distinct, then the positive number $\varepsilon = |l-k|/2$ is a valid error tolerance. [Convergence](/page/Convergence%20%28Real%20Sequences%29) to $l$ and convergence to $k$ then force a sufficiently late term $a_n$ of the [sequence](/page/Sequence) to be within $\varepsilon$ of both values. The triangle inequality then gives $|l-k| < |l-k|$, an impossibility. [/proofplan] [step:Assume the two proposed limits are distinct and choose half their distance] Suppose, for contradiction, that $l \ne k$. Define the [real number](/page/Real%20Numbers) $\varepsilon$ by \begin{align*} \varepsilon := \frac{|l-k|}{2}. \end{align*} Since $l \ne k$, the distance between $l$ and $k$ satisfies $|l-k| > 0$, so $\varepsilon > 0$. [/step] [step:Use both convergence hypotheses with the same tolerance] Since $a_n \to l$ as $n \to \infty$ and $\varepsilon > 0$, the definition of convergence gives a natural number $N_1 \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$ with $n \ge N_1$, \begin{align*} |a_n-l| < \varepsilon. \end{align*} Since $a_n \to k$ as $n \to \infty$ and the same $\varepsilon$ is positive, the definition of convergence gives a natural number $N_2 \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$ with $n \ge N_2$, \begin{align*} |a_n-k| < \varepsilon. \end{align*} Define $N := \max\{N_1,N_2\}$. Then for every $n \in \mathbb{N}$ with $n \ge N$, both inequalities hold. [guided] The reason for choosing the same tolerance for both limits is that we want a single term of the sequence to be close to both $l$ and $k$ at once. Since $a_n \to l$ as $n \to \infty$, the definition of convergence says that for every positive tolerance, and in particular for the positive number $\varepsilon$, there exists $N_1 \in \mathbb{N}$ such that \begin{align*} |a_n-l| < \varepsilon \end{align*} whenever $n \in \mathbb{N}$ and $n \ge N_1$. The second convergence hypothesis gives the same type of control around $k$. Since $a_n \to k$ as $n \to \infty$ and $\varepsilon > 0$, there exists $N_2 \in \mathbb{N}$ such that \begin{align*} |a_n-k| < \varepsilon \end{align*} whenever $n \in \mathbb{N}$ and $n \ge N_2$. To make both estimates valid simultaneously, define $N := \max\{N_1,N_2\}$. If $n \ge N$, then $n \ge N_1$ and $n \ge N_2$, so both \begin{align*} |a_n-l| < \varepsilon \qquad\text{and}\qquad |a_n-k| < \varepsilon \end{align*} hold for that same index $n$. [/guided] [/step] [step:Apply the triangle inequality to force an impossible strict inequality] Choose $n \in \mathbb{N}$ with $n \ge N$. Then \begin{align*} l-k = (l-a_n) + (a_n-k). \end{align*} Applying the triangle inequality in $\mathbb{R}$ gives \begin{align*} |l-k| &= |(l-a_n)+(a_n-k)| \\ &\le |l-a_n| + |a_n-k| \\ &= |a_n-l| + |a_n-k| \\ &< \varepsilon + \varepsilon \\ &= 2\varepsilon \\ &= |l-k|. \end{align*} Thus $|l-k| < |l-k|$, which is impossible for a real number. Therefore the assumption $l \ne k$ is false, and hence $l = k$. [/step]