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Created the Real Analysis Theorems and Definitions packet

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Alex Clarke
2024-01-09 13:01:34 -07:00
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chapter-4/chapter-4.tex
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\chapter{Limits}
\subimport{./}{limits-of-functions.tex}
\subimport{./}{limit-theorems.tex}
\subimport{./}{some-extensions-of-the-limit-concept.tex}
chapter-4/limit-theorems.tex
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\section{Limit Theorems}
\begin{definition}
Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. We say that $f$ is \textbf{bounded on a neighborhood of $c$} if there exists a $\delta$-neighborhood $V_\delta(c)$ of $c$ and a constant $M > 0$ such that we have $|f(x)| \leq M$ for all $x \in A \cap V_\delta (c)$.
\end{definition}
\begin{theorem}
If $A \subseteq \R$ and $f:A \rightarrow \R$ has a limit at $c \in \R$, then $f$ is bounded on some neighborhood of $c$.
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$ and let $f$ and $g$ be functions defined on $A$ to $\R$. We define the \textbf{sum} $f+g$, the \textbf{difference} $f-g$, and the \textbf{product} $fg$ on $A$ to $\R$ to be the functions given by
\[(f+g)(x):=f(x)+g(x),\]
\[(f-g)(x):=f(x)-g(x),\]
\[(fg)(x):=f(x)g(x)\]
for all $x \in A$. Further, if $b \in \R$, we define the \textbf{multiple} $bf$ to be the function given by
\[(bf)(x) := bf(x)\ \ \ \text{for all}\ \ \ x \in A\]
Finally, if $h(x)\neq 0$ for $x \in A$, we define the \textbf{quotient} $f/h$ to be the function given by
\[\left( \frac{f}{h}\right)(x) := \frac{f(x)}{h(x)}\ \ \ \ \text{for all}\ \ \ \ x \in A\]
\end{theorem}
\begin{theorem}
let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $c \in \R$ be a cluster point of $A$. Further, let $b \in \R$.
\begin{enumerate}
\item If $\lim\limits_{x\to c} f = L$ and $\lim\limits_{x\to c} g = M$, then
\[\lim\limits_{x\to c} (f+g) = L+M,\]
\[\lim\limits_{x\to c} (f-g)=L-M,\]
\[\lim\limits_{x\to c} (fg) = LM,\]
\[\lim\limits_{x\to c} (bf) = bL.\]
\item If $h: A \rightarrow \R$, if $h(x) \neq 0$ for all $x \in A$, and if $\lim\limits_{x\to c} h = H \neq 0$, then
\[\lim\limits_{x\to c} \left( \frac{f}{h} \right)= \frac{L}{H}\]
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f: A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. If
\[a \leq f(x) \leq b\ \ \ \ \text{for all}\ \ \ \ x \in A,\ x \neq c,\]
and if $\lim\limits_{x\to c} f$ exists, then $a \leq \lim\limits_{x\to c} f \leq b$.
\end{theorem}
\begin{theorem}[\textbf{Squeeze Theorem}]
Let $A \subseteq \R$, let $f,g,h:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. If
\[f(x) \leq g(x) \leq h(x)\ \ \ \ \text{for all}\ \ \ \ x \in A,\ x \neq c,\]
and if $\lim\limits_{x\to c} f = L = \lim\limits_{x\to c} h$, then $\lim\limits_{x\to c} g =L$.
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f:A \rightarrow \R$ and let $c \in \R$ be a cluster point of $A$. If
\[\lim\limits_{x\to c} f > 0\ \ \ \left[\textit{respectively, } \lim\limits_{x\to c} f < 0\right],\]
then there exists a neighborhood $V_\delta (c)$ of $c$ such that $f(x) > 0$ [respectively, $f(x) < 0$] for all $x \in A \cap V_\delta (c),\ x \neq c$.
\end{theorem}
chapter-4/limits-of-functions.tex
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\section{Limits of Functions}
\begin{definition}
Let $A \subseteq \R$. A point $c \in \R$ is a \textbf{cluster point} of $A$ if for every $\delta >0$ there exists at least one point $x \in A,\ x \neq c$ such that $|x-c|<\delta$.
\\\\This definition is rephrased in the language of neighborhoods as follows: A point $c$ is a cluster point of the set $A$ if every $\delta$-neighborhood $V_\delta (c)=(c-\delta, c+\delta)$ of $c$ contains at least one point of $A$ distinct from $c$.
\end{definition}
\begin{theorem}
A number $c \in \R$ is a cluster point of a subset $A$ of $\R$ if and only if there exists a sequence $(a_n)$ in $A$ such that $\lim (a_n) = c$ and $a_n \neq c$ for all $n \in \N$.
\end{theorem}
\begin{definition}
Let $A \subseteq \R$, and let $c$ be a cluster point of $A$. For a function $f:A \rightarrow \R$, a real number $L$ is said to be a \textbf{limit of $f$ at $c$} if, given any $\varepsilon>0$, there exists a $\delta>0$ such that if $x \in A$ and $0 < |x-c|<\delta$, then $|f(x)-L|<\varepsilon$.
\end{definition}
\begin{theorem}
If $f:A \rightarrow \R$ and if $c$ is a cluster point of $A$, then $f$ can have only one limit at $c$.
\end{theorem}
\begin{theorem}
Let $f:A \rightarrow \R$ and let $c$ be a cluster point of $A$. Then the following statements are equivalent.
\begin{enumerate}
\item $\lim\limits_{x\to c}=L$.
\item Given any $\varepsilon$-neighborhood $V_\varepsilon (L)$ of $L$, there exists a $\delta$-neighborhood $V_\delta (c)$ of $c$ such that if $x \neq c$ is any point in $V_\delta (c) \cap A$, then $f(x)$ belongs to $V_\varepsilon (L)$.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Sequential Criterion}]
Let $f:A \rightarrow \R$ and let $c$ be a cluster point of $A$. Then the following are equivalent.
\begin{enumerate}
\item $\lim\limits_{x\to c} f=L$.
\item For every sequence $(x_n)$ in $A$ that converges to $c$ such that $x_n\neq c$ for all $n \in \N$, the sequence $(f(x_n))$ converges to $L$.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Divergence Criteria}]
Let $A \subseteq \R$, let $f:A \rightarrow \R$ and let $c \in \R$ be a cluster point of $A$.
\begin{enumerate}
\item If $L \in \R$, then $f$ does \textbf{not} have limit $L$ at $c$ if and only if there exists a sequence $(x_n)$ in $A$ with $x_n \neq c$ for all $n \in \N$ such that the sequence $(x_n)$ converges to $c$ but the sequence $(f(x_n))$ does \textbf{not} converge to $L$.
\item The function $f$ does \textbf{not} have a limit at $c$ if and only if there exists a sequence $(x_n)$ in $A$ with $x_n \neq c$ for all $n \in \N$ such that the sequence $(x_n)$ converges to $c$ but the sequence $(f(x_n))$ does \textbf{not} converge in $\R$.
\end{enumerate}
\end{theorem}
Let the \textbf{signum function} sgn be defined by
\[\sign (x):=\begin{cases}
+1 & \text{for } x>0, \\
0 & \text{for } x=0, \\
-1 & \text{for } x < 0.
\end{cases}\]
chapter-4/some-extensions-of-the-limit-concept.tex
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\section{Some Extensions of the Limit Concept}
\begin{definition}
Let $A \in \R$ and let $f:A \rightarrow \R$.
\begin{enumerate}
\item If $c \in \R$ is a cluster point of the set $A \cap (c, \infty)= \{x \in A: x > c\}$, then we say that $L \in \R$ is a \textbf{right-hand limit of $f$ at $c$} and we write
\[\lim\limits_{x\to c^+} f=L\ \ \ \ \ \text{or}\ \ \ \ \ \lim\limits_{x\to c^+} f(x)=L\]
if given any $\varepsilon>0$ there exists a $\delta = \delta(\varepsilon)>0$ such that for all $x \in A$ with $0 < x-c < \delta$, then $|f(x)-L|<\varepsilon$.
\item If $c \in \R$ is a cluster point of the set $A \cap (-\infty, c)=\{x \in A: x <c\}$, then we say that $L \in \R$ is a \textbf{left-hand limit of $f$ at $c$} and we write
\[\lim\limits_{x\to c^-} f = L\ \ \ \ \ \text{or}\ \ \ \ \ \lim\limits_{x\to c^-} f(x)=L\]
if given any $\varepsilon > 0$ there exists a $\delta >0$ such that for all $x \in A$ with $0 < c-x < \delta$, then $|f(x)-L|<\varepsilon$.
\end{enumerate}
\end{definition}
\begin{theorem}
Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A \cap (c,\infty)$. Then the following statements are equivalent:
\begin{enumerate}
\item $\lim\limits_{x\to c^+} f = L$.
\item For every sequence $(x_n)$ that converges to $c$ such that $x_n \in A$ and $x_n > c$ for all $n \in \N$, the sequence $(f(x_n))$ converges to $L$.
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in \R$ be a cluster point of both of the sets $A \cap(c,\infty)$ and $A \cap (-\infty, c)$. Then $\lim\limits_{x\to c} f = L$ if and only if $\lim\limits_{x\to c^+} f = L = \lim\limits_{x\to c^-} f$.
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$.
\begin{enumerate}
\item We say that $f$ \textbf{tends to $\infty$ as $x \rightarrow c$}, and write
\[\lim\limits_{x\to c} f = \infty\]
if for every $\alpha \in \R$ there exists $\delta = \delta(\alpha) > 0$ such that for all $x \in A$ with $0 < |x-c|<\delta$, then $f(x) > \alpha$.
\item We say that $f$ \textbf{tends to $-\infty$ as $x \rightarrow c$}, and write
\[\lim\limits_{x\to c} f = - \infty\]
if for every $\beta \in \R$ there exists $\delta = \delta (\beta)>0$ such that for all $x \in A$ with $0 < |x-c|<\delta$, then $f(x)<\beta$.
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f,g:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. Suppose that $f(x) \leq g(x)$ for all $x \in A,\ x \neq c$.
\begin{enumerate}
\item If $\lim\limits_{x\to c} f = \infty$, then $\lim\limits_{x\to c} g = \infty$.
\item If $\lim\limits_{x\to c} g = -\infty$, then $\lim\limits_{x\to c} f = -\infty$.
\end{enumerate}
\end{theorem}
\begin{definition}
Let $A \subseteq \R$ and let $f:A \rightarrow \R$. If $c \in \R$ is a cluster point of the set $A \cap (c, \infty)= \{x \in A: x>c\}$, then we say that $f$ \textbf{tends to} $\infty$ [respectively, $-\infty$] as $x \rightarrow c^+$, and we write
\[\lim\limits_{x\to c^+} f = \infty\ \ \left[\text{respectively, } \lim\limits_{x\to c^+} f = -\infty\right]\]
if for every $\alpha \in \R$ there is $\delta = \delta(\alpha)>0$ such that for all $x \in A$ with $0 < x-c < \delta$, then $f(x)>\alpha$ [respectively, $f(x) < \alpha$]
\end{definition}
\begin{definition}
Let $A \subseteq \R$ and let $f:A \rightarrow \R$. Suppose that $(a,\infty) \subseteq A$ for some $a \in \R$. We say that $L \in \R$ is a \textbf{limit of $f$ as $x \rightarrow \infty$}, and write
\[\lim\limits_{x\to \infty} f = L\ \ \ \ \text{or}\ \ \ \ \lim\limits_{x\to \infty} f(x) = L,\]
if given any $\varepsilon > 0$ there exists $K=K(\varepsilon)>\alpha$ such that for any $x > K$, then $|f(x) - L| < \varepsilon$.
\end{definition}
\begin{theorem}
Let $A \subseteq \R$, let $f:A \rightarrow \R$, and suppose that $(a,\infty) \subseteq A$ for some $ a \in \R$. Then the following statements are equivalent:
\begin{enumerate}
\item $L=\lim\limits_{x\to \infty} f$.
\item For every sequence $(x_n)$ in $A \cap (a, \infty)$ such that $\lim (x_n) = \infty$, the sequence $(f(x_n))$ converges to $L$.
\end{enumerate}
\end{theorem}
\begin{definition}
let $A \subseteq \R$ and let $f: A \rightarrow \R$. Suppose that $(a,\infty)\subseteq A$ for some $a \in A$. We say that $f$ \textbf{tends to} $\infty$ [respectively, $ - \infty$] \textbf{as} $x \rightarrow \infty$, and write
\[\lim\limits_{x\to \infty} f = \infty \ \ \ \ \ \left[\text{respectively, } \lim\limits_{x\to \infty} f = - \infty \right]\]
if given any $\alpha \in \R$ there exists $K = K(\alpha)>\alpha$ such that for any $x > K$, then $f(x)>\alpha$ [respectively, $f(x) < \alpha$].
\end{definition}
\begin{theorem}
Let $A \in \R$, let $f:A \rightarrow \R$, and suppose that $(a, \infty) \subseteq A$ for some $a \in \R$. Then the following statements are equivalent:
\begin{enumerate}
\item $\lim\limits_{x\to \infty} = \infty$ [respectively, $\lim\limits_{x\to \infty} f = - \infty$]
\item For every sequence $(x_n)$ in $(a, \infty)$ such that $\lim (x_n) = \infty$, then $\lim (f(x_n)) = \infty$ [respectively, $\lim (f(x_n)) = - \infty$].
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $A \subseteq \R$, let $f,g:A \rightarrow \R$, and suppose that $(a,\infty) \subseteq A$ for some $a \in \R$. Suppose further that $g(x) > 0$ for all $x > a$ and that for some $L \in \R,\ L \neq 0$, we have
\[\lim\limits_{x\to \infty} \frac{f(x)}{g(x)} = L.\]
\begin{enumerate}
\item If $L > 0$, then $ \lim\limits_{x\to \infty} f = \infty$ if and only if $\lim\limits_{x\to \infty} g = \infty$.
\item If $L < 0$, then $\lim\limits_{x\to \infty} f = -\infty$ if and only if $\lim\limits_{x\to \infty} g = \infty$.
\end{enumerate}
\end{theorem}