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\subsubsection*{Limit of a function}

$f : E \subseteq \CC \to \CC$. We wish to define
what is meant by $\lim_{z \to a} f(z)$, even when
a might \emph{not} be in $E$. For example $\lim_{z
\to 0} \frac{\sin z}{z}$, with $E = \CC \setminus
\{0\}$. Also, if $E = \{0\} \cup [1, 2]$ it does
not make sense to speak about $z \in E$, $z \neq
0$, $z \to 0$.

\begin{definition*}
    $E \subseteq \CC$, $a \in \CC$. We say that
    $a$ is a \emph{limit point} of $E$ if for any
    $\delta > 0$, $\exists \ \in E$ such that $0 <
    |z - a| < \delta$.
\end{definition*}

\begin{remark*}
    $a$ is a limit point if and only if $\exists$
    a sequence $z_n \in E$ such that $z_n \to a$
    and $z_n \neq a \,\,\forall\,\,n$. (Check the
    equivalence!)
\end{remark*}

\begin{definition*}
    $f : E \subseteq \CC \to \CC$ and let $a \in
    \CC$ be a limit point of $E$. We say that
    $\lim_{z \to a} f(z) = l$ (``$f$ tends to $l$
    as $z$ tends to $a$'') if given $\eps > 0$,
    $\exists \delta > 0$ such that whenever $0 <
    |z - a| < \delta$ and $z \in E$, then $|f(z) -
    l| < \eps$.
\end{definition*}

\noindent
Equivalently: $f(z_n) \to l$ for every sequence
$z_i \in E$, $z_n \neq a$ and $z_n \to a$ (proved
exactly as last time with definition 1 $\iff$
definition 2).

\begin{remark*}
    Straight from the definitions we have that if
    $a \in E$ is a limit point, then $\lim_{z \to
    a} f(z) = f(a)$ if and only if $f$ is
    continuous at $a$.
\end{remark*}

\noindent
If $a \in E$ is \emph{isolated} (i.e. $a \in E$
and is not a limit point) then continuity of $f$
at $a$ always holds. \\
The limit of functions has very similar properties
to limit of sequences.
\begin{enumerate}[(1)]
    \item It is unique, $f(z) \to A$ and $f(z) \to
        B$ as $z \to a$
        \[ |A - B| \le |A - f(z)| + |f(z) - B| \]
        if $z \in E$ is such that $|z - a| <
        \delta_1, \delta_2$ then $|A - B| < 2\eps$
        so $A = B$. (the $\exists$ of such $z$ is
        consequence of the condition that $a$ is a
        limit point of $E$).
    \item $f(z) + g(z) \to A + B$ ($f(z) \to A$,
        $g(z) \to B$ as $z \to a$).
    \item $f(z)g(z) \to AB$
    \item if $B \neq 0$, $\frac{f(z)}{g(z)} \to
        \frac{A}{B}$ all proved in the same way as
        before.
\end{enumerate}

\subsection{The Intermediate Value Theorem}

\begin{theorem}
    $f : [a, b] \to \RR$ continuous and $f(a) \neq
    f(b)$. Then $f$ takes every value which lies
    between $f(a)$ and $f(b)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/cc50fb64974d11ec.png}
    \end{center}
    (for all $f(a) < \eta < f(b)$, $\exists c \in
    [a, b]$ such that $f(c) = \eta$)
\end{theorem}

\begin{proof}
    Without loss of generality we may suppose that
    $f(a) < f(b)$. Take $f(a) < \eta < f(b)$. Let
    \[ S = \{x \in [a, b] : f(x) < \eta\} \]
    $a \in S$, so $S \neq \emptyset$. Clearly $S$
    is bounded above by $b$. Then there is a
    supremum $c$ where $c \le b$. By definintion
    of supremum, given $n$, there exists $x_n \in
    S$ such that
    \[ c - \frac{1}{n} < x_n \le c \]
    so, $x_n \to c$ since $x_n \in S$, $f(x_n)
    < \eta$. By continuity of $f$, $f(x_n) \to
    f(c)$. Thus $f(c) \le \eta$. Now observe that
    $c \neq b$. Then for $n$ large, we can
    consider $c + \frac{1}{n} \in [a, b]$ and $c +
    \frac{1}{n} \to c$. Again by continuity
    \[ f \left( c + \frac{1}{n} \right) \to f(c) \]
    but since $c + \frac{1}{n} > c$, $f \left( c +
    \frac{1}{n} \right) \ge \eta$ (by definition
    of supremum). Hence $f(c) \ge \eta$ and
    therefore $f(c) = \eta$.
\end{proof}

\begin{remark*}
    The theorem is very useful for finding zeros
    or fixed points.
\end{remark*}

\begin{example*}
    Existence of the $n$-th root of a positive
    real number.
    \[ f(x) = x^n, \qquad x \ge 0 \]
    Let $y$ be a positive real number. $f$ is
    continuous on $[0, 1 + y]$ and
    \[ 0 = f(0) < y < (1 + y)^n = f(1 + y) \]
    so by the Intermediate Value Theorem, $\exists
    c \in (0, 1 + y)$ such that $f(c) = y$, i.e.
    $c^n = y$ so $c$ is a positive $n$-root of
    $y$. We also have uniqueness! (check)
\end{example*}
