% This is lecture no. 5 of MH3100/: Real Analysis I.

\chapter{Limit Theorems}%
\label{cha:limit_theorems}

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

\begin{thrm}
	Algebraic Limit Theorem.
	If a sequence $\left\{a_n\right\}$ converges to $a$ and a sequence $\left\{b_n\right\}$ converges to $b$, then
	\begin{enumerate}
		\item $\forall c\in\mathbb{R}$, the sequence $\left\{ca_n\right\}$ converges to $ca$;
		\item The sequence $\left\{a_n +b_n\right\}$ converges to $a+b$;
		\item The sequence $\left\{a_nb_n\right\}$ converges to $ab$; and
		\item The sequence $\ds{\left\{\frac{1}{a_n}\right\}}$ converges to $\ds{\frac{1}{a}}$, provided $a\neq 0$.
	\end{enumerate}
\end{thrm}

\begin{proof}
	$\lim ca_n = ca$.
	If $c\neq 0$, then the definition of $a=\lim a_n$ implies that
	\begin{align*}
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\backepsilon n>N & \implies |a_n -a|< \frac{\varepsilon}{|c|}\\
									      & \implies |ca_n -ca| <\varepsilon
	\end{align*}
	which is the definition of $ca=\lim(ca_n)$.
	If $c=0$, then $\left\{ca_n\right\}$ is a sequence of zeroes, which converges to $0=ca$.
\end{proof}
\begin{proof}
	$\lim a_n + b_n = a + b$.
	The definition of $a=\lim a_n$ and $b=\lim b_n$ implies that
	\begin{align*}
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\backepsilon n>N & \implies |a_n -a|, |b_n-b| < \frac{\varepsilon}{2}\\
									      &
		\begin{aligned}
			{} \implies |a_n +b_n -(a+b)| & \leqslant|a_n -a|+|b_n-b|\\
			                              & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}\\
						      & = \varepsilon,
		\end{aligned}
	\end{align*}
	which is the definition of $a + b=\lim(a_n + b_n)$.
\end{proof}
\begin{proof}
	$\lim a_nb_n = ab$.
	If $|a|+|b|>0$, then the definitions of $a=\lim a_n$ and $b=\lim b_n$ imply that
	\[
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\backepsilon n>N  \implies |a_n -a|, |b_n-b| < \min\left\{\sqrt{\frac{\varepsilon}{2}},\frac{\varepsilon}{2(|a|+|b|)}\right\}
	.\]
	And note that
	\begin{align*}
		\abs{a_nb_n - ab} & = \abs{a_nb_n - ab_n -a_nb + ab + ab_n -ab + a_nb -ab} \\
				  &\leqslant \abs{a_n-a}\abs{b_n-b} + \abs{a}\abs{b_n-b}+\abs{a_n-a}\abs{b}
	\end{align*}
	So by implication, we see that
	\begin{align*}
		\abs{a_nb_n-ab}   &\leqslant \abs{a_n-a}\abs{b_n-b} + \abs{a}\abs{b_n-b}+\abs{a_n-a}\abs{b}\\
				  & < \left(\sqrt{\frac{\varepsilon}{2}}\,\right)^2 + \frac{\abs{a}\times\varepsilon}{2(\abs{a}+\abs{b})} + \frac{\varepsilon\times\abs{b}}{2(\abs{a}+\abs{b})}\\
				  & = \varepsilon,
	\end{align*}
	which is the definition of $ab=\lim(a_nb_n)$.
	If however, $\abs{a}+\abs{b}=0$, then the definitions of $0=\lim a_n$ and $0=\lim b_n$ imply that
	\begin{align*}
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\backepsilon n>N & \implies |a_n|, |b_n| < \sqrt{\varepsilon}\\
									      &
		\begin{aligned}
		      {} \implies \abs{a_nb_n} & = \abs{a_n}\abs{b_n}\\
					    & < \sqrt{\varepsilon}\times\sqrt{\varepsilon}\\
					    & = \varepsilon,
		\end{aligned}
	\end{align*}
	which is the definition of $ab=0=\lim(a_nb_n)$.
\end{proof}
\begin{proof}
	$\lim \ds{\frac{1}{a_n}}=\ds{\frac{1}{a}}$.
	The definition of $a = \lim a_n$ implies that
	\begin{align*}
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\suchthat n>N & \implies\abs{a_n-a}<\frac{\abs{a}^2\varepsilon}{1+\abs{a}\varepsilon}\\
									   &
		\begin{aligned}
			{} \implies \abs*{\frac{1}{a_n}-\frac{1}{a}} & = \frac{\abs{a_n-a}}{\abs{a_n}\abs{a}}\\
								  & \leqslant \frac{\abs{a_n-a}}{(\abs{a}-\abs{a-a_n})\abs{a}}\\
								  & < \frac{\frac{\abs{a}^2\varepsilon}{1+\abs{a}\varepsilon}}{\left(\abs{a}-\frac{\abs{a}^2\varepsilon}{1+\abs{a}\varepsilon}\right)\abs{a}}\\
								  & = \varepsilon,
		\end{aligned}
	\end{align*}
	which is the definition of $\ds{\lim_{n \to \infty}\frac{1}{a_n}=\frac{1}{a}}$.
\end{proof}
\begin{thrm}
	Order Limit Theorem.
	For any convergent sequences $\left\{a_n\right\}$ and $\left\{b_n\right\}$,
	\begin{enumerate}
		\item if $\left\{a_n\right\}$ eventually satisfies $a_n\geqslant 0$, then $\lim a_n\geqslant 0$;
		\item if $\left\{a_n\right\}$ and $\left\{b_n\right\}$ eventually satisfy $a_n\leqslant b_n$, then $\lim a_n\leqslant b_n$; and
		\item $\forall c\in\mathbb{R}$, if $\left\{a_n\right\}$ eventually satisfies $a_n\leqslant c$, then $\lim a_n\leqslant c$.
	\end{enumerate}
\end{thrm}
\begin{proof}
	$a_n\geqslant 0\implies \lim a_n\geqslant 0$.
	Let $a$ denote the limit of $\left\{a_n\right\}$.
	For each $\varepsilon>0$, the definition of $a=\lim a_n$ implies that
	\begin{align*}
		\exists N\in\mathbb{N}\suchthat n>N & \implies a_n\geqslant 0, \abs{a_n-a}<\varepsilon\\
						    &
						    \begin{aligned}
							    {} \implies a & = (a-a_n)+a_n\\
								       & \geqslant a_n-\abs{a-a_n}\\
								       & > 0-\varepsilon\\
								       & = -\varepsilon
						    \end{aligned}\\
						    & \implies a > -\varepsilon.
	\end{align*}
	This shows that $a\geqslant 0$.
\end{proof}
\begin{proof}
	$a_n\leqslant b_n\implies \lim a_n\leqslant\lim b_n$.
	Since $\left\{b_n-a_n\right\}$ eventually satisfies $b_n-a_n\geqslant 0$, the first statement of this theorem asserts that $\lim(b_n-a_n)\geqslant 0$.
	Therefore, the Algebraic Limit Theorem implies that
	\begin{align*}
		\lim_{n \to \infty}b_n-\lim_{n \to \infty}a_n & =\lim_{n \to \infty}b_n + \lim_{n \to \infty}(-a_n)\\
							      & = \lim_{n \to \infty}(b_n-a_n)\\
							      & \geqslant 0.
	\end{align*}
	This shows that $\lim b_n\geqslant\lim a_n$.
\end{proof}
\begin{proof}
	$a_n\leqslant c\implies \lim a_n\leqslant c$.
	Since $\left\{a_n\right\}$ and the constant sequence $\left\{c\right\}$ eventually satisfies $a_n\leqslant c$, the second statement of the theorem asserts that
	\begin{align*}
		\lim_{n \to \infty}a_n & \leqslant \lim_{n \to \infty}c\\
				       & = c.
	\end{align*}
\end{proof}

\section{Squeeze Theorem}%
\label{sec:squeeze_theorem}

\begin{thrm}
	Squeeze Theorem.
	If the sequences $\left\{a_n\right\}$, $\left\{b_n\right\}$, and $\left\{c_n\right\}$ eventually satisfy
	\[
	a_n\leqslant b_n\leqslant c_n
	\]
	and $\left\{a_n\right\}$ and $\left\{c_n\right\}$ have the same limit $l$, then the sequence $\left\{b_n\right\}$ converges to $l$.
\end{thrm}
\begin{proof}
	Squeeze Theorem.
	The definition of $l=\lim a_n$ and $l=\lim c_n$ imply that
	\begin{align*}
		\forall\varepsilon>0,\ \exists N\in\mathbb{N}\suchthat n>N & \implies \abs{a_n-l},\abs{c_n-l}<\varepsilon\\
									   & \implies l-\varepsilon <a_n\leqslant b_n \leqslant c_n < l+\varepsilon\\
									   & \implies \abs{b_n-l}<\varepsilon,
	\end{align*}
	which is the definiton of $l=\lim b_n$.
\end{proof}
