September 17th, 2026

Updated 22 Sep 2026Notes (PDF)

[0.0.5]Definition(Sequence of Reals)#

A sequence is a function $f: \bZ_+ \to \bR$ . The $n$ -th term of $f$ is written as $x_n = f(n)$ . The entire sequence may be denoted as a list $(x_1, x_2, x_3, \dots)$ , or $(x_n)_{n = 1}^\infty$ .

[0.0.6]Definition(Limit of a Sequence)#

Let $(x_n)_{n = 1}^\infty \subseteq \bR$ be a sequence of reals. We say that the limit of the sequence $x_n$ is $L$ if, for every $\epsilon > 0$ , there exists some $N \in \bZ_+$ such that $|x_n - L| < \epsilon$ for every $n \ge N$ . We denote the limiting behavior in the following various forms.

\lim_{n \to \infty} x_n = L \quad \lim(x_n) = L \quad x_n \to L

[0.0.7]Theorem(Sum and Product of Limits)#

Let $(x_n)_{n = 1}^\infty$ and $(y_n)_{n = 1}^\infty$ be sequences of reals convering to $x$ and $y$ respectively.

$\lim(x_n + y_n) = x + y$ .
$\lim(x_ny_n) = xy$ .

Proof.

Fix $\epsilon > 0$ . There is some $N_1, N_2 \in \bZ_+$ such that $|x_n - x| < \epsilon/2$ and $|y_n - y| < \epsilon/2$ for every $n \ge N$ , where $N = \max\{N_1, N_2\}$ . Then note
$|(x_n + y_n) - (x + y)| \le |x_n - x| + |y_n - y| < \epsilon/2 + \epsilon/2 = \epsilon$
for every $n \ge N$ as desired.
Fix $\epsilon > 0$ . Since $x_n \to x$ , we have that there exists some $A \in \bZ_+$ such that $|x_n - x| < 1$ for every $n \ge A$ . We are justified in choosing
$M = \max\{|x_1|, |x_2|, \dots, |x_{A - 1}|, |x| + 1\},$
and note $|x_n| < M$ for all $n \in \bZ_+$ .
Let $N_1 \in \bZ_+$ be such that $|x_n - x| < \epsilon/2(|y| + 1)$ and $N_2 \in \bZ_+$ be such that $|y_n - y| < \epsilon/2M$ . Choose $N = \max\{N_1, N_2\}$ , and note
$|x_ny_n - xy| = |x_ny_n - yx_n + yx_n - xy| \le |x_n| |y_n - y| + |y| |x_n - x| < |x_n| \frac{\epsilon}{2M} + |y| \frac{\epsilon}{2(|y| + 1)} < \epsilon/2 + \epsilon / 2 = \epsilon$
for all $n \ge N$ . Thus $x_ny_n \to xy$ as desired.