% This is lecture no. 19 of MH3100/: Real Analysis I.

\chapter{Limit Theorems \& Divergence}%
\label{cha:limit_theorems_&_divergence}

\section{Limit Theorems}%
\label{sec:limit_theorems}

\begin{thrm}
	Algebraic Limit Theorem for Function Limits.
	If two functions $f,g\vcentcolon D\subseteq\mathbb{R}\mapsto\mathbb{R}$ have limits $L$ and $M$ respectively at a limit point $a$ of $D$, then
	\begin{enumerate}
		\item For all numbers $c\in\mathbb{R}$, the function $x\mapsto cf(x)$ has limit $cL$ at $a$;
		\item the function $x\mapsto f(x)+g(x)$ has limit $L+M$ at $a$;
		\item the function $x\mapsto f(x)g(x)$ has limit $LM$ at $a$; and
		\item the function $\displaystyle{x\mapsto \frac{1}{f(x)}}$ has limit $\displaystyle{\frac{1}{L}}$ at $a$, provided $L\neq 0$.
	\end{enumerate}
\end{thrm}

\begin{proof}
	$x\mapsto f(x)+g(x)$.
	The Sequential Criterion for Function Limits and the Algebraic Limit Theorem assert that
	\begin{align*}
		\forall\left\{x_n\right\}\in D\setminus\left\{a\right\}\quad a = \lim x_n & \implies f(x_n)\to L\text{ and }g(x_n)\to M\\
										          & \implies f(x_n) + g(x_n)\to L+M,
	\end{align*}
	which is the Sequential Criterion for $L+M=\lim_{x \to a}(f(x)+g(x))$.
\end{proof}

\begin{thrm}
	Order Limit Theorem for Function Limits.
	For any function $f,g\vcentcolon D\subseteq\mathbb{R}\mapsto\mathbb{R}$ with limits at a limit point $a$ of $D$,
	\begin{enumerate}
		\item if $f(x)\geqslant 0$ for all $x$ near $a$, then $\lim_{x \to a}f(x)\geqslant 0$;
		\item if $f(x)\leqslant g(x)$ for all $x$ near $a$, then $\lim_{x \to a}f(x)\leqslant \lim_{x \to a}g(x)$; and
		\item for all numbers $c\in\mathbb{R}$, if $f(x)\leqslant c$ for all $x$ near $a$, then $\lim_{x \to a}f(x)\leqslant c$.
	\end{enumerate}
\end{thrm}

\begin{proof}
	$f(x)\geqslant 0$.
	The Sequential Criterion for Limit Points asserts that
	\[
	\exists\left\{x_n\right\}\in D\setminus\left\{a\right\}\suchthat a=\lim x_n
	.\]
	The Sequential Criterion for Function Limits then asserts
	\[
	\lim_{x \to a}f(x)=\lim f(x_n)
	.\]
	Since $x_n\to a$ and satisfies $x_n\neq a$, $\left\{x_n\right\}$ is eventually arbitrarily near $a$.
	Therefore, $\left\{f(x_n)\right\}$ eventuallys satisfies $f(x_n)\geqslant 0$.
	The Order Limit Theorem then asserts $\lim f(x_n)\geqslant 0$.
\end{proof}

\begin{thrm}
	Squeeze Theorem for Function Limits.
	If the functions $f,g,h\vcentcolon D\subseteq\mathbb{R}\to\mathbb{R}$ satisfy
	\[
	f(x)\leqslant g(x)\leqslant h(x)
	\]
	for all $x$ near a limit point $a$ of $D$, and $f$ and $h$ have the same limit $L$ at $a$, then the function $g$ has limit $L$ at $a$.
\end{thrm}

\begin{proof}
	$\forall\left\{x_n\right\}\in D\setminus\left\{a\right\}$:
	\begin{align*}
		a=\lim x_n & \implies \left\{x_n\right\}\text{ is eventually arbitrarily near }a\\
			   & \implies  \left\{f(x_n)\right\},\left\{g(x_n)\right\},\left\{h(x_n)\right\}\\
			   & \hphantom{\implies}\mathrel{} \text{eventually satisfy }f(x_n)\leqslant g(x_n)\leqslant h(x_n).
	\end{align*}
	Moreover, the Sequential Criterion for Function Limits asserts that
	\[
		\forall\left\{x_n\right\}\in D\setminus\left\{a\right\}\quad a=\lim x_n\implies f(x_n)\to L\text{ and }h(x_n)\to L
	.\]
	Therefore, the Squeeze Theorem asserts that
	\[
	\forall\left\{x_n\right\}\in D\setminus\left\{a\right\}\quad a=\lim x_n\implies g(x_n)\to L
	\]
	which is the Sequential Criterion for $L=\lim_{x \to a}g(x)$.
\end{proof}

\section{Divergence Criterion}%
\label{sec:divergence_criterion}

\begin{thrm}
	First Divergence Criterion for Function Limits.
	A function $f\vcentcolon D\subseteq\mathbb{R}\mapsto\mathbb{R}$ \textbf{does not have a limit} at a limit point $a$ of $D$ iff there is a sequence $\left\{x_n\right\}$ \textbf{in} $D\setminus\left\{a\right\}$ \textbf{converging to} $a$ for which the sequence $\left\{f(x_n)\right\}$ \textbf{is divergent}.
\end{thrm}

\begin{proof}
	The Sequential Criterion for Function Limits asserts that
	\[
	\forall\left\{x_n\right\}\in D\setminus\left\{a\right\}\quad a=\lim x_n\implies \lim_{x \to a}f(x)=\lim f(x_n)
	.\]
	The Sequential Criterion for Limit Points asserts that
	\[
		\exists\left\{x_n\right\}\in D\setminus\left\{a\right\}\text{ with }a=\lim x_n
	.\]
	The Sequential Criterion for Function Limits then asserts that
	\begin{align*}
		& \exists\left\{y_n\right\}\in D\setminus\left\{a\right\}\text{ with }a=\lim y_n\text{ and }\\
		& \left\{f(x_n)\right\},\left\{f(y_n)\right\}\text{ do not have the same limit. }
	\end{align*}
	Then $\left\{z_n\right\}\defas x_1,y_1,x_2,y_2,x_3,y_3,\dotsc\in D\setminus\left\{a\right\}$ converges to $a$ for which the Limits of Subsequences asserts that $\left\{f(z_n)\right\}$ is divergent.
\end{proof}

\begin{thrm}
	Second Divergence Criterion for Function Limits.
	A function $f\vcentcolon D\subseteq\mathbb{R}\mapsto\mathbb{R}$ that is \textbf{bounded} near a limit point $a$ of $D$ \textbf{does not have a limit at} $a$ iff there are two sequences $\left\{x_n\right\}$, $\left\{y_n\right\}$ in $D\setminus\left\{a\right\}$ \textbf{converging to} $a$ for which the sequences $\left\{f(x_n)\right\}$, $\left\{f(y_n)\right\}$ \textbf{have different limits}.
\end{thrm}
