\section{Limits}



\definition{Boundedness}
Let $a_1, a_2, a_3, ... $ (abbreviated $a_n$ in this handout) be a sequence of real numbers. \par

We say a number $u$ is an \textit{upper bound} of $a_n$ if $a_i \leq u ~ \forall i$. \par
If $a_n$ has an upper bound, we could say it is \textit{bounded from above}.



\definition{Monotonically increasing}
We say a sequence of real numbers $a_n$ is \textit{monotonically increasing} if $m < n \implies a_m < a_n$.


\definition{Limits (informal)}
The \textit{limit} of a sequence is a point to which the sequence gets \say{closer} to. \par
Not all sequences have limits. \par
Don't let this hand-wavy definition bother you too much, we'll see a proper definition soon.


\theorem{}<limexists>
Any monotonically increasing sequence that is bounded from above has a unique limit. \par

\vfill



\problem{}
Let $a_1 = 2$, and $a_{n+1} = 2 + \sqrt{a_n}$. \par

Show that the sequence $a_n$ is monotonically increasing and bounded above. Find its limit.

\begin{solution}
	\textbf{$a_n$ is monotonically increasing:} Show this by induction. \par
	\textbf{$a_n$ is bounded:} Induction again. Show that $\sqrt{a_n} < 2$. \par

	\textbf{$\lim_{n \to\infty}a_n = 4$:} Use the fact that $\lim_{n\to\infty} a_n = \lim_{n\to\infty}a_{n+1}$, the hit the resulting equation with some algebra. Note that $1$ cannot be a solution, since $a_n \geq 2\ \forall n$.


\end{solution}

\vfill



\problem{}
Show that both assumptions of \ref{limexists} are necessary: \par

Find an example of a monotonically increasing sequence that does not have a limit, \par
and of a bounded sequence that does not have a limit.

\vfill
\pagebreak



\definition{Limits (formal)}
Let $a_n$ be a sequence.
$L$ is the \textit{limit} of this sequence if for any $\varepsilon < 0$, \par
we can find an $N$ so that $|a_n - L| < \varepsilon \forall n \geq N$.

\vfill



\problem{}
Show that $0$ is the limit of $a_n = \frac{1}{n}$, where $n \geq 1$. \par
Show that $\pi$ is the limit of the sequence $3,~ 3.1,~ 3.14,~ ...$.

\begin{solution}
	$\lim_{n\to\infty}a_n = 0$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n| < \epsilon\ \forall n < N$. \par

	We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \par
	It is now clear that $\lim_{n\to\infty}a_n = 0$

	\linehack{}

	$\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \par

	Looking at the definition of this sequence, we get a similar ``Archemedian Property'': \par
	$\forall \epsilon$, $\exists m$ so that $|a_m - \pi| < \epsilon$.
\end{solution}

\vfill




\problem{}
Show that if a sequence $a_n$ has a limit, that limit is unique. \par
\hint{Show that if $A, B$ are both limits of $a_n$, $A$ and $B$ must be equal.}

\begin{solution}
	If both $A$ and $B$ are limits of $[a_n]$, we have the following: \par
	$\forall \epsilon > 0$, $\exists N_A \in \mathbb{N}$ so that $|a_n - A| < \epsilon\ \forall n < N_A$. \par
	$\forall \epsilon > 0$, $\exists N_B \in \mathbb{N}$ so that $|a_n - B| < \epsilon\ \forall n < N_B$. \par

	Let $N = \max(N_A, N_B)$. \par
	Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \par
	which can be writen as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \par

	By the triangle inequality, we have \par
	$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \par
	And since $|a_n - A + B - a_n| = |B - A|$, we have \par
	$|B - A| < 2\epsilon\ \forall n > N$. \par

	This should be true for all $\epsilon > 0$. \par
	Let's set $\epsilon = \frac{|B-A|}{4}$, which is greater than zero iff $A \neq B$ \par
	Then, $|B-A| < \frac{|B-A|}{2}$, which is impossible\textsuperscript{*}. \par

	Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal.

	\linehack{}

	\textsuperscript{*}Note that we can also set $\epsilon = \frac{|B-A|}{2}$, which gives $|B-A| < |B-A|$. This is also false, and the proof still works. However, the extra $\frac{\hspace{2em}}{2}$ gives a clearer explanation.

\end{solution}

\vfill
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