The Timpson Lyceum · Real Analysis (2020)

Lecture 6 (Real Analysis 2020)

Overview

This lecture builds the algebra of limits for sequences — the collection of theorems that let us compute limits of sums, scalar multiples, products, and quotients without returning to the $\varepsilon$ – $N$ definition every time. We first reinforce direct $\varepsilon$ – $N$ verification (using the Archimedean property) and prove that a sequence enumerating $\mathbb{Q}$ fails to converge. Then we assemble the limit laws, prove that a nonzero limit forces the sequence to stay bounded away from zero, and use that to handle reciprocals and quotients. Everything culminates in the polynomial limit theorem: the limit of a polynomial evaluated along a convergent sequence equals the polynomial evaluated at the limit, proved by induction on the degree.

Key Concepts

Direct $\varepsilon$ – $N$ verification. To prove $\lim_{n\to\infty} a_n = L$ directly means: given any $\varepsilon > 0$ , exhibit an index $N$ such that $|a_n - L| < \varepsilon$ for all $n \ge N$ — using only the definition (and the Archimedean property), not previously proved limit theorems.

Archimedean property: for any real $c > 0$ there exists $N \in \mathbb{N}$ with $\tfrac{1}{N} < c$ . This is the engine that turns an inequality in $\varepsilon$ into a concrete index.
Bounding strategy: in scratch work, simplify the quantity $|a_n - L|$ by replacing pieces with larger quantities (never smaller). If the enlarged expression is $< \varepsilon$ , so is the original.
Monotonicity of the denominator: for $n \ge N$ , $\tfrac{1}{\sqrt{n}} \le \tfrac{1}{\sqrt{N}}$ — a bigger denominator gives a smaller fraction.

Showing a sequence does NOT converge. A sequence that hits two fixed values infinitely often (here, $1$ and $\tfrac12$ ) cannot converge.

If $a_n \to a$ , then for $\varepsilon = \tfrac14$ there is an index $N$ with $|a_n - a| < \tfrac14$ for all $n \ge N$ ; i.e. eventually all terms lie in an interval of radius $\tfrac14$ about $a$ .
No single point $a$ can be within $\tfrac14$ of both $1$ and $\tfrac12$ (their midpoint $\tfrac34$ sits exactly $\tfrac14$ from each, so neither is strictly inside). Producing arbitrarily late indices $n_1, n_2 \ge N$ with $a_{n_1} = \tfrac12$ and $a_{n_2} = 1$ yields a contradiction.

The limit laws (algebra of limits). Suppose $\lim a_n = a$ and $\lim b_n = b$ .

Sum: $\lim (a_n + b_n) = a + b$ .
Scalar multiple: for $\alpha \in \mathbb{R}$ , $\lim (\alpha\, a_n) = \alpha\, a$ .
Product: $\lim (a_n b_n) = a\,b$ .
Reciprocal: if $b \ne 0$ , $\lim \tfrac{1}{b_n} = \tfrac{1}{b}$ .
Quotient: if $b \ne 0$ , $\lim \tfrac{a_n}{b_n} = \tfrac{a}{b}$ .
Linearity: $\lim (\alpha\, a_n + \beta\, b_n) = \alpha\, a + \beta\, b$ for $\alpha,\beta \in \mathbb{R}$ — a single statement packaging the sum and scalar rules. An operation is linear when it distributes over scalar combinations this way.

Comparison limit (tool from the previous lecture). If $\lim a_n = a$ and there exist $c > 0$ and an index $N_1$ with $|b_n - b| \le c\,|a_n - a|$ for all $n \ge N_1$ , then $\lim b_n = b$ . This lets us bound an unknown limit by a known one.

Bounded-away-from-zero lemma. If $\lim b_n = b$ with $b \ne 0$ , then there is an index $N$ such that $|b_n| > \tfrac{|b|}{2} \quad \text{for all } n \ge N.$ Apply the definition with $\varepsilon = \tfrac{|b|}{2}$ : for $n \ge N$ , $|b_n - b| < \tfrac{|b|}{2}$ forces $b_n$ onto the same side of $0$ as $b$ and keeps it past $\tfrac{|b|}{2}$ in magnitude. This guards against dividing by terms that creep toward $0$ , and is the key step in the reciprocal law.

Bounded sequences (preview of the next section). A sequence $\{a_n\}$ is bounded if there exists $M$ with $|a_n| \le |M|$ (equivalently $|a_n| \le M$ for some $M \ge 0$ ) for all $n$ — it runs off to neither $+\infty$ nor $-\infty$ . A sequence is contained in a set $S$ when every term $a_n \in S$ . A set $S$ is dense in $\mathbb{R}$ iff every $x \in \mathbb{R}$ is the limit of some sequence lying in $S$ .

Worked Example

Reciprocal law. Show that if $\lim b_n = b$ with $b \ne 0$ , then $\lim \tfrac{1}{b_n} = \tfrac{1}{b}$ .

By the lemma, fix $N$ so that $|b_n| > \tfrac{|b|}{2}$ for all $n \ge N$ ; equivalently $\tfrac{1}{|b_n|} < \tfrac{2}{|b|}$ . For such $n$ , bound the difference and then invoke the comparison limit against $b_n \to b$ :

\begin{aligned} \left|\frac{1}{b_n} - \frac{1}{b}\right| &= \left|\frac{b - b_n}{b\,b_n}\right| = \frac{1}{|b|}\cdot\frac{1}{|b_n|}\cdot |b_n - b| \\ &\le \frac{1}{|b|}\cdot\frac{2}{|b|}\cdot |b_n - b| = \frac{2}{|b|^2}\,|b_n - b| \qquad (n \ge N). \end{aligned}

Since $|b_n - b| \to 0$ and the multiplier $c = \tfrac{2}{|b|^{2}}$ is a fixed positive constant, the comparison limit gives $\lim \tfrac{1}{b_n} = \tfrac{1}{b}$ .

Quotient as a corollary. Once the reciprocal law and product law are in hand, the quotient law is immediate:

\begin{aligned} \lim \frac{a_n}{b_n} &= \lim\!\left(a_n \cdot \frac{1}{b_n}\right) = \big(\lim a_n\big)\!\left(\lim \frac{1}{b_n}\right) \\ &= a \cdot \frac{1}{b} = \frac{a}{b} \qquad (b \ne 0). \end{aligned}

Polynomial limit theorem (induction on degree). For a polynomial $p(x) = \sum_{k=0}^{m} a_k x^k$ and a sequence $b_j \to b$ , claim $\lim p(b_j) = p(b)$ .

Base case ( $m = 1$ ): $p_1(b_j) = a_0 + a_1 b_j$ , so by the sum and scalar laws $\lim (a_0 + a_1 b_j) = a_0 + a_1 b = p_1(b)$ .
Inductive step: assume $\lim p_k(b_j) = p_k(b)$ for every degree- $k$ polynomial. Split off the top term of a degree- $(k{+}1)$ polynomial:

\begin{aligned} \lim p_{k+1}(b_j) &= \lim\!\Big( p_k(b_j) + a_{k+1}\,b_j^{\,k+1} \Big) \\ &= \lim p_k(b_j) + a_{k+1}\,\lim b_j^{\,k+1} = p_k(b) + a_{k+1}\,b^{\,k+1} = p_{k+1}(b), \end{aligned}

using the sum and scalar laws, the inductive hypothesis, and repeated application of the product law to get $\lim b_j^{\,k+1} = b^{\,k+1}$ . By induction the result holds for every degree.

Summary

A direct limit proof produces, for each $\varepsilon > 0$ , an explicit $N$ via the Archimedean property; simplify $|a_n - L|$ only by enlarging it.
A sequence that attains two distinct values infinitely often (e.g. $1$ and $\tfrac12$ ) cannot converge — no center $a$ keeps both within $\tfrac14$ .
The limit laws — sum, scalar multiple, product, reciprocal, quotient, and the packaged linearity rule — let you evaluate limits compositionally instead of from the definition.
If $\lim b_n = b \ne 0$ , then eventually $|b_n| > \tfrac{|b|}{2}$ ; this bounded-away-from-zero fact is what makes the reciprocal and quotient laws work.
The polynomial limit theorem — $\lim p(b_j) = p(b)$ — follows by induction on degree from the sum, scalar, and product laws; you may "pass the limit inside" any polynomial.
Looking ahead: a sequence is bounded when $|a_n| \le M$ for some $M$ , and a set is dense in $\mathbb{R}$ when every real is the limit of a sequence drawn from it.