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# Chapter 4 - Limits
## Secrion 4.1 - Limits of Functions
## Section 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 $0 < |a - c| < \delta$.
**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 that $a \neq c$. 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 $a_n \rightarrow c$
**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood conttains infinitely many points of $A$.
**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood contains infinitely many points of $A$.
**Definition**. The set of every cluster point of $A$ is called the *derived set* of $A$, and denoted $A'$.
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---
**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
**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 given 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**. 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**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set with 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**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$, there exists some $V_\delta(c)$ such that for 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 $x \in A$,
@@ -78,4 +78,4 @@ $$
f(x) \leq g(x) \leq h(x) \; \text{ for all } x \in A, x \neq c
$$
Then, $\lim_{x \rightarrow c} g(x) = L$.
Then, $\lim_{x \rightarrow c} g(x) = L$.