82 lines
3.8 KiB
Markdown
82 lines
3.8 KiB
Markdown
# Chapter 4 - Limits
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## Section 4.1 - Limits of Functions
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**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 $0 < |a - c| < \delta$.
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**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 $a_n \rightarrow c$
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**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 $A$.
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**Definition**. The set of every cluster point of $A$ is called the *derived set* of $A$, and denoted $A'$.
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**Corollary**. A set $A$ is closed if and only if $A' \subseteq A$.
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**Remark**. If $A'$ is the derived set of $A$, then $A'' \subseteq A'$.
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**Remark**. Intervals involving infinity and square brackets for the constant are closed.
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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 given any $\varepsilon > 0$, there exists some $\delta > 0$ such that
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$$
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0 < |x-c| < \delta \Rightarrow |f(x) - L| < \varepsilon
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$$
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**Theorem**. For a given function and cluster point, there can be at most one limit at said point.
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**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 $L$.
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---
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**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.
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**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 $x \in V_\varepsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
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**Definition**. The limit of a function at infinity is defined if for a given $\varepsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,
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$$
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x > \alpha \Rightarrow |f(x) - L| < \varepsilon
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$$
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## Section 4.2 - Limit Theorems
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**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 $|f(x)| \leq M$.
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**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 $c$.
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**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:
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$$\lim_{x \rightarrow c} (f(x) + g(x)) = L + M$$
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$$\lim_{x \rightarrow c} (f(x)g(x)) = LM$$
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Additionally, if $g(x) \neq 0$ for all $x \in A$, and $M \neq 0$, then
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$$
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\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{L}{M}
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$$
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**Corollary**. If $p, q \in \mathbb{R}[x]$, and $q(c) \neq 0$ for some $c \in \mathbb{R}$, then
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$$
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\lim_{x \rightarrow c} p(x) = p(c)
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$$
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$$
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\lim_{x \rightarrow c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}
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$$
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**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
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$$
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\lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} h(x) = L
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$$
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$$
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f(x) \leq g(x) \leq h(x) \; \text{ for all } x \in A, x \neq c
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$$
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Then, $\lim_{x \rightarrow c} g(x) = L$.
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