Finding an $N$ for one $\varepsilon$, as in the definition, is a computation. A proof must handle every $\varepsilon > 0$ at once, so $N$ becomes a formula in $\varepsilon$. Every ε–N proof is written in two passes: scratch work backward, then a clean argument forward.
🔑 Key idea
Scratch work (not part of the proof): simplify $|a_n - L|$, bound it by something simple like $\frac{C}{n}$, and solve $\frac{C}{n} < \varepsilon$ to see that $N > \frac{C}{\varepsilon}$ will do. The proof: "Let $\varepsilon > 0$. Choose an integer $N > \frac{C}{\varepsilon}$. Let $n \ge N$. Then $|a_n - L| \le \frac{C}{n} \le \frac{C}{N} < \varepsilon$."
💡 Why it works
The order of the sentences follows the order of the quantifiers in the definition: for every $\varepsilon$ (let $\varepsilon > 0$), there is an $N$ (choose it, using only $\varepsilon$), such that for all $n \ge N$ (let $n \ge N$), the inequality holds (the estimate). The scratch work runs the other way, starting from the inequality we want, which is why it cannot be handed in as a proof: it assumes what it is trying to show. Replacing $|a_n - L|$ by a simpler upper bound is allowed and makes the algebra easy; the bound only has to be true and to go to $0$.
Model proof. Claim: $\frac{3n + 1}{n + 2} \to 3$. Scratch: $\left|\frac{3n + 1}{n + 2} - 3\right| = \frac{5}{n + 2} < \frac{5}{n}$, and $\frac{5}{n} < \varepsilon$ when $n > \frac{5}{\varepsilon}$. Proof. Let $\varepsilon > 0$. By the Archimedean property choose an integer $N > \frac{5}{\varepsilon}$. For every $n \ge N$, $\left|\frac{3n + 1}{n + 2} - 3\right| = \frac{5}{n + 2} < \frac{5}{n} \le \frac{5}{N} < \varepsilon$. $\blacksquare$
The proof's choice $N > \frac{5}{\varepsilon}$ for $\frac{3n + 1}{n + 2} \to 3$: try several $\varepsilon$ and check that $N = \frac{5}{\varepsilon}$ (rounded up) always lands in the green.
Worked examples
✏️ Worked example: Simplify, then bound
For $a_n = \frac{2n - 7}{n + 4}$ and $L = 2$, find a constant $C$ with $|a_n - L| < \frac{C}{n}$, and write the choice of $N$.
Handing in the scratch work. "Suppose $\frac{5}{n + 2} < \varepsilon$; then $n > \frac{5}{\varepsilon} - 2$, so the limit is $3$" starts from the conclusion. Another common slip is an $N$ that depends on $n$ (like "choose $N = n + 1$"): $N$ must be fixed before $n$ is chosen, from $\varepsilon$ alone. And a proof that picks $\varepsilon = 0.01$ proves the inequality for that one $\varepsilon$, not convergence.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.