3.8 KiB
Chapter 4 - Limits
Secrion 4.1 - Limits of Functions
Definition. Let A \subseteq \mathbb{R}. Then, a point c \in \mathbb{R} is a cluster point of A if for every \delta > 0, the $\delta$-neighborhood of c contains a point a \in A such thhat a \neq c. That is, there exists some a such that 0 < |a - c| < \delta.
Theorem. A real number c is a cluster point for a set A if and only if there exists a sequence (a_n) in A\\ \{c\} such that a_n \rightarrow c
Collary. A real number c is a cluster point of a set A if and only if every $\delta$-neighborhood conttains infinitely many points of A.
Definition. The set of every cluster point of A is called the derived set of A, and denoted A'.
Collary. A set A is closed if and only if A' \subseteq A.
Remark. If A' is the derived set of A, then A'' \subseteq A'.
Remark. Intervals involving infinity and square brackets for the constant are closed.
Definition. Suppose f: A \rightarrow \mathbb{R} is a function with domain A \subseteq \mathbb{R}, and let c \in A be a cluster point of A. then, a real number L is a limit of f at $c$ if goven any \epsilon > 0, there exists some \delta > 0 such that
0 < |x-c| < \delta \Rightarrow |f(x) - L| < \epsilon
Therorem. For a given function and cluster point, there can be at most one limit at said point.
Theorem. Let A \subseteq \mathbb{R} and f: A \rightarrow \mathbb{R}. Then, to show that lim_{x \rightarrow c} f(x) = L, it suffices to show that for every sequence (a_n) in A\\ \{c\}, the sequence (f(a_n)) converges tto L.
Definition. The extended real numbers are \hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \} are a totally-ordered set witth supremum and infimum. Note that this set is no longer a field.
Definition. At any point c, the limitt of f at c is infinite if given some \alpha, there exists some V_\delta(c) such that forr all x \in V_\epsilon(c), then f(x) \in V_\alpha(\infty).
Definition. The limit of a function at infinity is defined if for a given \epsilon, there exists some \alpha so that there exists some V_\delta(c) such that for all x \in A,
x > \alpha \Rightarrow |f(x) - L| < \epsilon
Section 4.2 - Limit Theorems
Definition. Let A \subseteq \mathbb{R} and c \in \mathbb{R} be a cluster point of A. Then, a function f: A \rightarrow \mathbb{R} is bounded on a neighborhood of $c$ if there exists some $\delta$-neighborhood V_\delta(c) of c and some constant M > 0 such that for all x \in A \cap V_\delta(c), then |f(x)| \leq M.
Theorem. If A \subseteq \mathbb{R} and f: A \rightarrow \mathbb{R} has a finite limit at c \in \mathbb{R}, then f is bounded on some neighborhood of c.
Theorem. With A \subseteq \mathbb{R}, and f, g: A \rightarrow \mathbb{R}, with c \in \mathbb{R} a cluster point of A, then if \lim_{x \rightarrow c} f(x) = L and \lim_{x \rightarrow c} g(x) = M, then:
\lim_{x \rightarrow c} (f(x) + g(x)) = L + M
\lim_{x \rightarrow c} (f(x)g(x)) = LM
Additionally, if g(x) \neq 0 for all x \in A, and M \neq 0, then
\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{M}
Collary. If p, q \in \mathbb{R}[x], and q(c) \neq 0 for some c \in \mathbb{R}, then
\lim_{x \rightarrow c} p(x) = p(c)
\lim_{x \rightarrow c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}
Theorem. Squeeze Theorem. Let A \subseteq \mathbb{R}. Then, if f, g, h: A \rightarrow \mathbb{R} and with c \in \mathbb{R} being a cluster point of A, then if both
\lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} h(x) = L
f(x) \leq g(x) \leq h(x) \; \text{ for all } x \in A, x \neq c
Then, \lim_{x \rightarrow c} g(x) = L.