In Calculus 1 you used the limit laws freely. Now we prove them from the definition, which shows how ε-proofs combine. You will need boundedness for the product.
🔑 Key idea
If $a_n \to a$ and $b_n \to b$, then $a_n + b_n \to a + b$, $c\,a_n \to ca$, $a_n b_n \to ab$, and $\frac{a_n}{b_n} \to \frac{a}{b}$ when $b \ne 0$. The proofs split $\varepsilon$ between the pieces ("$\frac{\varepsilon}{2}$ each") and use $N = \max(N_1, N_2)$ so that both estimates hold at once.
💡 Why it works
Sum. Let $\varepsilon > 0$. Choose $N_1$ with $|a_n - a| < \frac{\varepsilon}{2}$ for $n \ge N_1$, and $N_2$ with $|b_n - b| < \frac{\varepsilon}{2}$ for $n \ge N_2$. For $n \ge N = \max(N_1, N_2)$, the triangle inequality gives $|(a_n + b_n) - (a + b)| \le |a_n - a| + |b_n - b| < \varepsilon$. $\blacksquare$ Product. Add and subtract $a_n b$: $$a_n b_n - ab = a_n(b_n - b) + b(a_n - a).$$ Since $(a_n)$ converges it is bounded, say $|a_n| \le M$ with $M > 0$. Choose $N$ so that $|b_n - b| < \frac{\varepsilon}{2M}$ and $|a_n - a| < \frac{\varepsilon}{2|b| + 1}$ for $n \ge N$. Then $|a_n b_n - ab| < M\frac{\varepsilon}{2M} + |b|\frac{\varepsilon}{2|b| + 1} < \varepsilon$. $\blacksquare$ (The $+1$ avoids dividing by $0$ when $b = 0$.)
Quotient. It is enough to show $\frac{1}{b_n} \to \frac{1}{b}$, then use the product law. Since $b_n \to b \ne 0$, eventually $|b_n - b| < \frac{|b|}{2}$, and then $|b_n| > \frac{|b|}{2}$ by the reverse triangle inequality. So $\left|\frac{1}{b_n} - \frac{1}{b}\right| = \frac{|b - b_n|}{|b_n||b|} < \frac{2}{b^{2}}|b_n - b|$, which can be made smaller than any $\varepsilon$.
Worked examples
✏️ Worked example: Splitting $\varepsilon$ with numbers
Suppose $|a_n - a| \le \frac{3}{n}$ and $|b_n - b| \le \frac{5}{n}$. For $\varepsilon = 0.1$, find $N_1$, $N_2$ and $N$ as in the sum proof.
- Each piece must be less than $\frac{\varepsilon}{2} = 0.05$.
- $\frac{3}{n} < 0.05$ means $n > 60$, so $N_1 = 61$; $\frac{5}{n} < 0.05$ means $n > 100$, so $N_2 = 101$.
Answer: $N_1 = 61$, $N_2 = 101$, $N = 101$.
✏️ Worked example: Using the laws
If $a_n \to 3$ and $b_n \to -2$, find $\lim \frac{a_n b_n - a_n^{2}}{b_n + 4}$.
- The denominator tends to $2 \ne 0$, so the quotient law applies.
- The numerator tends to $3(-2) - 9 = -15$.
Answer: $-\frac{15}{2}$
✏️ Worked example: The step that protects a denominator
Suppose $b_n \to -4$. Explain why eventually $|b_n| > 2$, and bound $\left|\frac{1}{b_n} - \left(-\frac{1}{4}\right)\right|$ in terms of $|b_n + 4|$.
- With $\varepsilon = 2 = \frac{|b|}{2}$ there is an $N$ with $|b_n + 4| < 2$ for $n \ge N$. By the reverse triangle inequality, $|b_n| \ge 4 - |b_n + 4| > 2$.
- Then $\left|\frac{1}{b_n} + \frac{1}{4}\right| = \frac{|b_n + 4|}{4|b_n|} < \frac{|b_n + 4|}{8}$.
Answer: For $n \ge N$: $\left|\frac{1}{b_n} + \frac{1}{4}\right| < \frac{1}{8}|b_n + 4|$, which goes to $0$.
⚠️ Common mistake
Using one $N$ for two sequences. "Choose $N$ so that $|a_n - a| < \frac{\varepsilon}{2}$ and $|b_n - b| < \frac{\varepsilon}{2}$" is fine only after saying why one $N$ serves both: take the larger of the two. In the product proof, the slip is forgetting to bound $|a_n|$: the factor $a_n$ changes with $n$, so "$|a_n||b_n - b|$ is small" needs $|a_n| \le M$ first.