Limits and Continuity

Introduction

In this post, I will discuss limits and continuity. There are 3 purposes for writing this post:

1. To practise my academic writing.
2. To write for fun.
3. To expound some points that I was stuck on for a while at first when I was learning as an undergraduate in my first semester. It may help someone, or not, because the main purposes actually are 1&2.

I hope you can enjoy the read.

Limits of Sequences

First, I would like to note that all the discussions will be based on $\mathbb{R}$, and the discussions will not involve metric spaces. Before we continue, we need to state two axioms.

Axiom 1 (Archimedes’ axiom)
For every $a, b > 0$, there exists $N \in \mathbb{N}$, such that $b < Na$.

Axiom 2 (Cantor’s axiom)
${a_n}$ and ${b_n}$ are two given sequences satisfying ${a_n} < {b_n}$ for every $n$, with associated intervals $[a_n,b_n]$ nested and their lengths tending to zero. Then there exists a unique $c$ such that $\bigcap\limits_{n=1}^{\infty} [a_n, b_n]=\{c\}$.

These axioms will be used later.

1.1Definition For every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \ge N$, $|a_n - \alpha| < \varepsilon$. We write $a_n \to \alpha$ as $n \to \infty$, or $$\lim\limits_{n \to \infty}a_n = \alpha$$

We call $\alpha$ limit value and we say the limit of sequence $\{a_n\}$ is $\alpha$ or $\{a_n\}$ converges to $\alpha$

Remark. If for every $M > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \ge N, a_n \ge M$, then it means that the sequence $\{a_n\}$ does not converge(it diverges). Furthermore, the sequence diverges to positive infinity(for every $M < 0$, $a_n \le M$, we say the sequence diverges to negative infinity). We write: $$\lim_{n \to \infty} a_n = +\infty \qquad a_n \rightarrow +\infty \ (n \to \infty)$$

or

$$\lim_{n \to \infty} a_n = -\infty \qquad a_n \rightarrow -\infty \ (n \to \infty)$$

1.2 Theorem If $\{a_n\}$ is convergent, the limit is unique.

Proof. Suppose that there are two different limit values, $\alpha$ and $\beta$. Since $\alpha \neq \beta$, we let $\varepsilon = \frac{|\alpha - \beta|}{2}$. By 1.1 Definition, there exists an $N \in \mathbb{N}$, such that for all $n \ge N$, $|a_n - \alpha| < \varepsilon$ and $|a_n - \beta| < \varepsilon$ (taking the larger $N$ if necessary). Then we have $|\alpha - \beta| = |(a_n - \alpha) - (a_n - \beta)| \le |a_n - \alpha| + |a_n - \beta| < 2\varepsilon = |\alpha - \beta|$. This is a contradiction.

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1.3 Theorem (Squeeze Theorem) Let $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ be sequences of real numbers. Suppose that $\{a_n\}$ and $\{b_n\}$ converge to $L$, and also suppose that for every $n$, we have $a_n < c_n < b_n$. Then $\lim\limits_{n \to \infty} c_n = L$.

Proof. Since $\{a_n\}$ and $\{b_n\}$ converge to $L$, it follows that for every given $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \ge N$, $|a_n - L| < \varepsilon$ and $|b_n - L| < \varepsilon$. Moreover, since for every $n$, $a_n < c_n < b_n$, we obtain $a_n - L < c_n - L < b_n - L \Rightarrow -\varepsilon < c_n - L < \varepsilon \Rightarrow |c_n - L| < \varepsilon$. Hence $\lim\limits_{n \to \infty} c_n = L$.

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1.4 Theorem

1. $\lim\limits_{n \to \infty}(a_n + b_n) = \alpha + \beta$
2. $\lim\limits_{n \to \infty}(a_n - b_n) = \alpha - \beta$
3. $\lim\limits_{n \to \infty}(c a_n) = c \alpha$
4. $\lim\limits_{n \to \infty}(a_n b_n) = \alpha \beta$
5. $\lim\limits_{n \to \infty}(\frac{a_n}{b_n}) = \frac{\alpha}{\beta} \quad (\lim\limits_{n \to \infty} b_n \neq 0, \beta \neq 0)$

Only the first one will be proved; the others can be exercises for practice.

Proof. Let $\varepsilon > 0$ be given. Because $a_n$ and $b_n$ are convergent, it follows that for $\varepsilon' = \frac{\varepsilon}{2}$, there exists an $N_1$ such that for all $n > N_1$, $|a_n - \alpha| < \varepsilon'$. $b_n$ is the same: we know there exists an $N_2$. Let $N= \max(N_1, N_2)$. It is obvious that for every $n > N$, both $|a_n - \alpha| < \varepsilon'$ and $|b_n - \beta| < \varepsilon'$ are true. Now we estimate: $|(a_n + b_n) - (\alpha + \beta)| = |a_n - \alpha + b_n - \beta| \le |a_n - \alpha| + |b_n - \beta| < \varepsilon' + \varepsilon' = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$. Hence, by 1.1 Definition, $\lim_{n\to\infty}(a_n+b_n) = \alpha+\beta$.

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