[proofplan] We prove each property of limits of complex sequences directly from the $\epsilon$-$N$ definition. Uniqueness follows from the triangle inequality applied to two putative limits. The algebraic rules (sum, product, reciprocal) use the triangle inequality together with boundedness of convergent sequences. Bound preservation follows by contradiction, reducing the lower bound case to the upper bound case via negation. [/proofplan] [step:Establish uniqueness of limits via the triangle inequality] Assume $a_n \to a$ and $a_n \to b$. Let $\epsilon > 0$ be given. By [convergence](/pages/1211) to $a$, there exists $N_1 \in \mathbb{N}$ such that $|a_n - a| < \epsilon/2$ for all $n \geq N_1$. By convergence to $b$, there exists $N_2 \in \mathbb{N}$ such that $|a_n - b| < \epsilon/2$ for all $n \geq N_2$. Set $N = \max(N_1, N_2)$. For $n \geq N$, the triangle inequality gives \begin{align*} |a - b| \leq |a - a_n| + |a_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Since $\epsilon > 0$ is arbitrary, $|a - b| = 0$, hence $a = b$. [guided] We want to show that if a [sequence](/pages/1149) converges, it converges to exactly one [limit](/pages/1146). The strategy is to assume two limits $a$ and $b$ exist, then show $|a - b|$ is smaller than every positive number, forcing $a = b$. Let $\epsilon > 0$ be given. By the definition of $a_n \to a$, there exists $N_1 \in \mathbb{N}$ such that $|a_n - a| < \epsilon/2$ for all $n \geq N_1$. Similarly, by $a_n \to b$, there exists $N_2 \in \mathbb{N}$ such that $|a_n - b| < \epsilon/2$ for all $n \geq N_2$. Set $N = \max(N_1, N_2)$. The key idea is that for any $n \geq N$, the term $a_n$ is simultaneously close to both $a$ and $b$, so $a$ and $b$ must be close to each other. By the triangle inequality: \begin{align*} |a - b| \leq |a - a_n| + |a_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Since $\epsilon > 0$ was arbitrary and $|a - b| \geq 0$ is a fixed non-negative real number less than every positive number, we conclude $|a - b| = 0$, hence $a = b$. [/guided] [/step] [step:Show that subsequences of a convergent sequence converge to the same limit] Assume $a_n \to a$ and let $(n_j)_{j=1}^{\infty}$ be a strictly increasing sequence of natural numbers. Let $\epsilon > 0$ be given. There exists $N \in \mathbb{N}$ such that $|a_n - a| < \epsilon$ for all $n \geq N$. Since $(n_j)$ is strictly increasing, an induction argument shows $n_j \geq j$ for all $j \geq 1$. Therefore for $j \geq N$, we have $n_j \geq j \geq N$, so \begin{align*} |a_{n_j} - a| < \epsilon. \end{align*} Hence $a_{n_j} \to a$. [/step] [step:Verify that constant sequences converge to their constant value] Let $a_n = c$ for all $n$. For any $\epsilon > 0$ and any $n \geq 1$: \begin{align*} |a_n - c| = |c - c| = 0 < \epsilon. \end{align*} Taking $N = 1$ satisfies the definition, so $a_n \to c$. [/step] [step:Prove the sum rule using the triangle inequality with $\epsilon/2$ splitting] Assume $a_n \to a$ and $b_n \to b$. Let $\epsilon > 0$ be given. There exists $N_1 \in \mathbb{N}$ with $|a_n - a| < \epsilon/2$ for all $n \geq N_1$, and $N_2 \in \mathbb{N}$ with $|b_n - b| < \epsilon/2$ for all $n \geq N_2$. Set $N = \max(N_1, N_2)$. For $n \geq N$, the triangle inequality gives \begin{align*} |(a_n + b_n) - (a + b)| &\leq |a_n - a| + |b_n - b| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Therefore $a_n + b_n \to a + b$. [guided] The idea is to split the combined error $(a_n + b_n) - (a + b) = (a_n - a) + (b_n - b)$ and control each piece separately. Let $\epsilon > 0$ be given. We need each piece to contribute at most $\epsilon/2$ to the total error. Since $a_n \to a$, there exists $N_1 \in \mathbb{N}$ with $|a_n - a| < \epsilon/2$ for all $n \geq N_1$. Since $b_n \to b$, there exists $N_2 \in \mathbb{N}$ with $|b_n - b| < \epsilon/2$ for all $n \geq N_2$. Set $N = \max(N_1, N_2)$ so that both estimates hold simultaneously for $n \geq N$. By the triangle inequality: \begin{align*} |(a_n + b_n) - (a + b)| &= |(a_n - a) + (b_n - b)| \\ &\leq |a_n - a| + |b_n - b| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Therefore $a_n + b_n \to a + b$. The $\epsilon/2$-splitting is the standard technique for combining two independent convergence estimates. [/guided] [/step] [step:Prove the product rule by splitting the error and using boundedness] Assume $a_n \to a$ and $b_n \to b$. Since convergent [sequences](/pages/1149) are bounded, there exists $M > 0$ such that $|a_n| \leq M$ and $|b| \leq M$ for all $n$. Let $\epsilon > 0$ be given. There exists $N_1 \in \mathbb{N}$ with $|a_n - a| < \epsilon/(2M)$ for all $n \geq N_1$, and $N_2 \in \mathbb{N}$ with $|b_n - b| < \epsilon/(2M)$ for all $n \geq N_2$. Set $N = \max(N_1, N_2)$. For $n \geq N$, add and subtract $a_n b$: \begin{align*} |a_n b_n - ab| &= |a_n b_n - a_n b + a_n b - ab| \\ &\leq |a_n||b_n - b| + |b||a_n - a| \\ &\leq M \cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M} = \epsilon. \end{align*} Therefore $a_n b_n \to ab$. [guided] The product $a_n b_n - ab$ does not split as neatly as a sum, so we introduce the intermediate term $a_n b$. The algebraic identity is: \begin{align*} a_n b_n - ab = a_n(b_n - b) + b(a_n - a). \end{align*} The first term involves $a_n$, which varies with $n$ -- we need it bounded. Since $(a_n)$ converges, it is bounded: there exists $M > 0$ with $|a_n| \leq M$ for all $n$. We also need $|b|$ bounded, which holds with $|b| \leq M$ (enlarging $M$ if necessary). Now we choose $N_1, N_2$ so that $|a_n - a| < \epsilon/(2M)$ for $n \geq N_1$ and $|b_n - b| < \epsilon/(2M)$ for $n \geq N_2$. Setting $N = \max(N_1, N_2)$, the triangle inequality gives \begin{align*} |a_n b_n - ab| &\leq |a_n||b_n - b| + |b||a_n - a| \\ &\leq M \cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M} = \epsilon. \end{align*} Therefore $a_n b_n \to ab$. The key insight is that boundedness of the convergent factor $a_n$ converts a product estimate into a linear one. [/guided] [/step] [step:Prove the reciprocal rule by first establishing a lower bound on $|a_n|$] Assume $a_n \to a$ with $a_n \neq 0$ for all $n$ and $a \neq 0$. Since $|a| > 0$, there exists $N_0 \in \mathbb{N}$ such that $|a_n - a| < |a|/2$ for all $n \geq N_0$. The [reverse triangle inequality](/theorems/2300) gives \begin{align*} |a_n| \geq |a| - |a_n - a| > |a| - \frac{|a|}{2} = \frac{|a|}{2} \end{align*} for all $n \geq N_0$. Now let $\epsilon > 0$ be given. There exists $N_1 \in \mathbb{N}$ such that $|a_n - a| < |a|^2 \epsilon / 2$ for all $n \geq N_1$. Set $N = \max(N_0, N_1)$. For $n \geq N$: \begin{align*} \left|\frac{1}{a_n} - \frac{1}{a}\right| &= \frac{|a - a_n|}{|a||a_n|} \\ &< \frac{|a - a_n|}{|a| \cdot |a|/2} \\ &= \frac{2|a - a_n|}{|a|^2} \\ &< \frac{2}{|a|^2} \cdot \frac{|a|^2 \epsilon}{2} = \epsilon. \end{align*} Therefore $1/a_n \to 1/a$. [guided] The difficulty in the reciprocal rule is that $1/a_n$ can blow up if $|a_n|$ is small. We need a uniform lower bound on $|a_n|$ for large $n$. Since $a_n \to a$ and $a \neq 0$, eventually $a_n$ stays within distance $|a|/2$ of $a$. Specifically, there exists $N_0 \in \mathbb{N}$ such that $|a_n - a| < |a|/2$ for all $n \geq N_0$. The reverse triangle inequality then gives \begin{align*} |a_n| \geq |a| - |a_n - a| > |a| - \frac{|a|}{2} = \frac{|a|}{2} \end{align*} for all $n \geq N_0$. This is the crucial bound: it says $|a_n|$ is eventually bounded away from zero. Now we estimate the difference of reciprocals. For $n \geq N_0$: \begin{align*} \left|\frac{1}{a_n} - \frac{1}{a}\right| = \frac{|a - a_n|}{|a| \cdot |a_n|} < \frac{|a - a_n|}{|a| \cdot |a|/2} = \frac{2|a - a_n|}{|a|^2}. \end{align*} To make this less than $\epsilon$, we need $|a - a_n| < |a|^2 \epsilon / 2$. Choose $N_1 \in \mathbb{N}$ with $|a_n - a| < |a|^2 \epsilon / 2$ for all $n \geq N_1$, and set $N = \max(N_0, N_1)$. Then for $n \geq N$: \begin{align*} \left|\frac{1}{a_n} - \frac{1}{a}\right| < \frac{2}{|a|^2} \cdot \frac{|a|^2 \epsilon}{2} = \epsilon. \end{align*} Therefore $1/a_n \to 1/a$. [/guided] [/step] [step:Prove upper bound preservation by contradiction] Assume $a_n \leq A$ for all $n$ and $a_n \to a$. Define $t_n = A - a_n \geq 0$ for each $n$. By the sum rule (applied to the constant sequence $A$ and $-a_n$), $t_n \to A - a$. Suppose for contradiction that $a > A$. Set $\delta = a - A > 0$. Since $t_n \to A - a = -\delta$, there exists $N \in \mathbb{N}$ such that $|t_n - (A - a)| < \delta/2$ for all $n \geq N$. This gives \begin{align*} t_n < (A - a) + \frac{\delta}{2} = -\delta + \frac{\delta}{2} = -\frac{\delta}{2} < 0, \end{align*} contradicting $t_n \geq 0$. Therefore $a \leq A$. [/step] [step:Reduce lower bound preservation to the upper bound case via negation] Assume $A \leq a_n$ for all $n$ and $a_n \to a$. Define $b_n = -a_n$ and $B = -A$. Then $b_n \leq B$ for all $n$, and the sum rule gives $b_n \to -a$. By the upper bound preservation result, $-a \leq B = -A$, which rearranges to $A \leq a$. [/step]