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Finished section sets and functions

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chapters/sections/lims_and_funcs.tex
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@ -49,4 +49,580 @@ In this chapter we will introduce the notation
x \in \realn \text{ is a boundary point of } D \subset \realn \iff \exists \anyseqdef{D} \text{ such that } x_n \rightarrow x
\]
\end{thm}
\begin{proof}
Let $x$ be a boundary point of $D$. Then
\begin{equation}
\forall n \in \natn ~\exists x_n \in D \cap \left( x - \frac{1}{n}, x+ \frac{1}{n} \right)
\end{equation}
The resulting sequence $\seq{x}$ is in $D$, and
\begin{equation}
|x - x_n| \le \frac{1}{n}
\end{equation}
holds. Therefore, $x_n$ converges to $x$. Now let $\anyseqdef{D}$, with $x_n \rightarrow x$. This means
\begin{equation}
\forall \epsilon > 0 ~\exists N \in \natn: ~~|x - x_N| < \epsilon
\end{equation}
Then
\begin{equation}
x_N \in D \cap \oball(x)
\end{equation}
Since $\epsilon$ is arbitrary, $x$ is a boundary point of $D$.
\end{proof}
\begin{defi}
Let $D \subset \realn$ and $f: D \rightarrow \realn$. Let $x_0$ be a boundary point of $D$.
We say that $f$ converges to $y \in \realn$ for $x \rightarrow x_0$ and write
\[
\lim_{x \rightarrow x_0} f(x) = y
\]
if
\[
\forall \epsilon > 0 ~\exists \delta > 0: ~~|x-x_0| < \delta \implies |f(x) - f(y)| < \epsilon
\]
\end{defi}
\begin{rem}
This definition only makes sense for boundary points $x_0$ of $D$. The most imoprtant case is
\[
D = (x_0 - a, x_0 + a) \setminus \set{x_0}
\]
\end{rem}
\begin{eg}
\begin{enumerate}[(i)]
\item Let $a \in \realn$
\begin{align*}
f: \realn &\longrightarrow \realn \\
x &\longmapsto ax
\end{align*}
Consider $a \ne 0$: Let $\epsilon > 0$. We want that
\[
|f(x) - 0| = |a||x| \stackrel{!}{<} \epsilon
\]
Choose $\delta = \frac{\epsilon}{|a|}$. Then we have
\[
|x| = |x - 0| < \delta \implies |f(x) - 0| = |a||x| < |a| \delta = |a| \frac{\epsilon}{|a|} = \epsilon
\]
Therefore
\[
\lim_{x \rightarrow 0} f(x) = 0
\]
\item Consider
\begin{align*}
f: \realn &\longrightarrow \realn \\
x &\longmapsto \begin{cases}
1 ,& x > 0 \\
-1 ,& x < 0 \\
\end{cases}
\end{align*}
$f$ doesn't converge for $x \rightarrow 0$. Assume $y \in \realn$ is the limit of $x$ at $0$. This means that there is a $\delta > 0$ such that
\[
|f(x) - y| < 1 \text{ if } |x| = |x - 0| < \delta
\]
Then, for any $x \in (0, \delta)$ we have
\[
2 = |f(x) - f(-x)| \le \underbrace{|f(x) - y|}_{< 1} + \underbrace{|y - f(-x)|}_{< 1} < 2
\]
which is a contradiction.
\end{enumerate}
\end{eg}
\begin{thm}
Let $f: D \rightarrow \realn$, $x_0$ a boundary point of $D$ and $y \in \realn$. Then
\[
\lim_{x \rightarrow x_0} f(x) = y \iff \forall \anyseqdef{D} \text{ with } x_n \longrightarrow x_0: ~~\limn f(x_n) = x_0
\]
\end{thm}
\begin{proof}
Assume that $\lim_{x \rightarrow x_0} f(x)$ and that there is $\anyseqdef{D}$ converging to $x$.
Let $\epsilon > 0$, then
\begin{equation}
\exists \delta > 0: ~~|x - x_0| < \delta \implies |f(x) - y| < \epsilon
\end{equation}
Since $x_n \rightarrow x_0$, we know that
\begin{equation}
\exists N \in \natn ~\forall n > N: ~~|x_n - x_0| < \delta
\end{equation}
For such $n$, the epsilon criterion $|f(x_n) - y| < \epsilon$ also holds, and thus
\begin{equation}
f(x_n) \convinf y
\end{equation}
Now to prove the "$\impliedby$" direction, assume that $\lim_{x \rightarrow x_0} f(x) \ne y$, i.e.
\begin{equation}
\exists \epsilon > 0 ~\forall \delta > 0 ~\exists x \in D: ~~|x - x_0| < \delta \wedge |f(x) - y| \ge \epsilon
\end{equation}
Choose $\forall x \in \natn$ one $x_n$ such that
\begin{equation}
|x_n - x_0| < \frac{1}{n} \text{ but } |f(x_n) - y| \ge \epsilon
\end{equation}
Then $x_n \rightarrow x_0$, but $|f(x_n) - y| \ge \epsilon ~\forall n \in \natn$, so
\begin{equation}
\limn f(x_n) \ne y
\end{equation}
This indirectly proves "$\impliedby$".
\end{proof}
\begin{eg}
Consider $D = \realn \subset \set{0}$, we want to prove
\[
\lim_{x \rightarrow 0} \frac{1}{1-x} = 1
\]
So let $\anyseqdef{D}$ with $x_n \rightarrow 0$. Then
\[
\frac{1}{1-x_n} \convinf 1
\]
\[
\implies \lim_{x \rightarrow 0} \frac{1}{1-x} = 1
\]
However, the limit $\lim_{x \rightarrow 1}$ doesn't exist. Let $x_n = \frac{1}{n} + 1$ with $x_n \rightarrow 1$. Then
\[
\frac{1}{1 - (\frac{1}{n} + 1)} = -n \convinf -\infty
\]
This doesn't converge, thus there is no limit.
\end{eg}
\begin{cor}
Let $f, g: D \rightarrow \realn$, $x_0$ a boundary point and $y, z \in \realn$ such that
\begin{align*}
\lim_{x \rightarrow x_0} f(x) = y && \lim_{x \rightarrow x_0} g(x) = z
\end{align*}
Then
\begin{align*}
\lim_{x \rightarrow x_0} (f(x) + g(x)) &= y+z \\
\lim_{x \rightarrow x_0} (f(x) \cdot g(x)) &= y \cdot z
\end{align*}
If $z \ne 0$, then
\[
\lim_{x \rightarrow x_0} (\frac{f(x)}{g(x)}) = \frac{y}{z}
\]
\end{cor}
\begin{proof}
Here we will only prove the last statement.
Let $\lim_{x \rightarrow x_0} = z \ne 0$. Then
\begin{equation}
\exists \delta > 0 ~\forall x \in \oball[\delta](x_0): ~~|g(x) - z| < |z|
\end{equation}
$g$ doesn't have any roots on $\oball[\delta](x_0)$. Let $\anyseqdef{D \cap \oball[\delta](x_0)}$ converge to $x_0$.
According to prerequisites, we have
\noindent\begin{subequations}
\begin{multicols}{2}
\noindent
\begin{equation}
\limn f(x_n) = y
\end{equation}
\begin{equation}
\limn g(x_n) = z \ne 0
\end{equation}
\end{multicols}
\end{subequations}
\noindent Thus
\begin{equation}
\implies \limn \frac{f(x_n)}{g(x_n)} = \frac{y}{z} \implies \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = \frac{y}{z}
\end{equation}
\end{proof}
\begin{cor}[Squeeze Theorem]
Let $f, g, h: D \rightarrow \realn$ and $x$ a boundary point of $D$. If for $y \in \realn$
\[
\lim_{x \rightarrow x_0} f(x) = y = \lim_{x \rightarrow x_0} h(x)
\]
and
\[
f(x) \le g(x) \le h(x) ~~\forall x \in \oball(x_0)
\]
then
\[
\lim_{x \rightarrow x_0} g(x) = y
\]
\end{cor}
\begin{eg}
Consider $\exp(x)$. We already know that
\[
1 + x \le \exp(x) ~~\forall x \in \realn
\]
This also implies that
\[
1 - x \le \exp(-x) = \frac{1}{\exp(x)} ~~\forall x \in \realn
\]
So
\[
1 + x \le \exp(x) \le \frac{1}{1 - x}
\]
The limits of these terms are
\begin{align*}
\lim_{x \rightarrow 0} (1+x) = 1 && \lim_{x \rightarrow 0} \left(\frac{1}{1-x}\right) = 1
\end{align*}
And using the squeeze theorem this results in
\[
\lim_{x \rightarrow 0} \exp(0) = 1
\]
\end{eg}
\begin{defi}
Let $f: D \rightarrow \realn$ and $x_0$ a boundary point of $D$. We say $f$ diverges to infinity for $x \rightarrow x_0$ and write
\[
\lim_{x \rightarrow x_0} f(x) = \infty
\]
if
\[
\forall K \in (0, \infty) ~\exists \delta > 0: ~~|x - x_0| < \delta \implies f(x) \ge K
\]
\end{defi}
\begin{defi}
Let $D \subset \realn$ be unbounded above. We say $f$ converges for $x \rightarrow \infty$ to $y \in \realn$ and write
\[
\lim_{x \rightarrow \infty} f(x) = y
\]
if
\[
\forall \epsilon > 0 ~\exists K \in (0, \infty) ~\forall x > K: ~~|f(x) - y| < \epsilon
\]
\end{defi}
\begin{rem}
Let $f: D \rightarrow \cmpln$ and $x_0$ a boundary point of $D$. Then
\begin{align*}
&\limes{x}{x_0} f(x) = y \in \cmpln \\
\iff & \limes{x}{x_0} \Re(f(x)) = \Re(y) \wedge \limes{x}{x_0} \Im(f(x)) = \Im(y) \\
\iff & \limes{x}{x_0} |f(x) - y| = 0
\end{align*}
\end{rem}
\begin{defi}
Let $D \subset K$, $f:D \rightarrow K$ and $x_0 \in D$. $f$ is called continuous in $x_0$ if
\[
\forall \epsilon > 0 ~\exists \delta > 0: ~~|x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon
\]
If $f$ is continuous in every point of $D$, we call $f$ continuous.
$f$ is called Lipschitz
continuous if
\[
\exists L \in (0, \infty) ~\forall x, y \in D: ~~|f(x) - f(y)| \le L|x - y|
\]
$L$ is called Lipschitz constant
\end{defi}
\begin{rem}
Let $f: D \rightarrow \field$
\[
f \text{ is continuous in } x_0 \in D \iff \limes{x}{x_0} f(x) = f(x_0)
\]
\end{rem}
\begin{eg}
We want to show that
\begin{align*}
f: \realn &\longrightarrow \realn \\
x &\longmapsto x^2
\end{align*}
is continuous. To do that, let $x_0 \in \realn$, $\epsilon > 0$. We want
\[
|f(x) - f(x_0)| = |x^2 - x_0^2| = |x - x_0||x + x_0| \must[\le] \epsilon
\]
So we choose
\[
\delta = \min \set{1, \frac{\epsilon}{2|x_0| + 1}} > 0
\]
Then for every $x$ with $|x - x_0| < \delta$
\begin{align*}
|f(x) - f(x_0)| &= |x - x_0||x + x_0| \le \delta(|x| + |x_0|) \le \delta(|x_0| + \delta + |x_0|) \\
&\le \delta(2|x_0| + 1) \le \frac{\epsilon}{2|x_0| + 1}(2|x_0| + 1) = \epsilon
\end{align*}
\end{eg}
\begin{thm}
Every Lipschitz continuous function is continuous
\end{thm}
\begin{proof}
Let $f: D \rightarrow \field$ be a Lipschitz continuous function with Lipschitz constant $L > 0$. I.e.
\begin{equation}
\forall x, y \in D: ~~|f(x) - f(y)| \le L|x - y|
\end{equation}
Let $x_0 \in \realn$ and $\epsilon > 0$. Choose $\delta = \frac{\epsilon}{L}$. Then $|x - x_0| < \delta$ implies
\begin{equation}
|f(x) - f(x_0)| \le L|x - x_0| \le L \cdot \delta = \epsilon
\end{equation}
\end{proof}
\begin{eg}
\begin{enumerate}[(i)]
\item Consider the constant function $x \mapsto c$, $c \in \field$.
\[
|f(x) - f(y)| = |c - c| = 0 \le 1 \cdot |x - y|
\]
\item Consider the linear function $x \mapsto cx$, $c \in \field$.
\[
|f(x) - f(y)| = |cx - cy| = |c||x - y|
\]
\end{enumerate}
These two functions are Lipschitz continuous, and therefore continuous.
\begin{enumerate}[(i)]\setcounter{enumi}{2}
\item Consider $x \mapsto \Re(x)$. Then
\[
|\Re(x) - \Re(y)| = |\Re(x - y)| \le |x - y|
\]
Analogously this works for $\Im(x)$. Both of those are Lipschitz continuous.
\item Lipschitz continuity depends on $D$. E.g.
\begin{align*}
f: [0, 1] &\longrightarrow \realn \\
x &\longmapsto x^2
\end{align*}
is Lipschitz continuous:
\[
|f(x) - f(y)| = |x-y||x+y| \le 2 \cdot |x - y|
\]
However,
\begin{align*}
g: \realn &\longrightarrow \realn \\
x &\longmapsto x^2
\end{align*}
is NOT Lipschitz continuous, because
\[
\frac{|g(n+1) - g(n)|}{(n+1) - n} = 2n + 1 \convinf \infty
\]
\end{enumerate}
\end{eg}
\begin{rem}
Let $f: D \rightarrow \field$.
\begin{gather*}
f \text{ is continuous in } x_0 \in D \\
\iff \\
\forall \anyseqdef{D} \text{ with } x_n \rightarrow x_0: ~~\limn f(x_n) = f(x_0)
\end{gather*}
If $f, g$ are continuous in $x_0$, then $f+g$ and $f \cdot g$ are also continuous in $x_0$, and if $g(x_0) \ne 0$ then $f/g$ is also continuous in $x_0$.
Notably, polynomials are continuous. A rational function (the quotient of two polynomials) is continuous in all points that are not roots of the denominator.
\end{rem}
\begin{thm}
Let $D \subset \field$, and let
\begin{subequations}
\begin{gather}
f: D \longrightarrow \field \text{ continuous in } x_0 \in D \\
g: f(D) \longrightarrow \field \text{ continuous in } f(x_0)
\end{gather}
\end{subequations}
Then $g \circ f$ is also continuous in $x_0$.
\end{thm}
\begin{proof}
Let $\epsilon > 0$. Since $g$ is continuous in $f(x_0)$,
\begin{equation}
\exists \delta_1 > 0: ~~|y - f(x_0)| < \delta_1 \implies |g(y) - g(f(x_0))| < \epsilon
\end{equation}
Since $f$ is continuous in $x_0$,
\begin{equation}
\exists \delta_2 > 0: ~~|x - x_0| < \delta_2 \implies |f(x) - f(x_0)| < \delta_1
\end{equation}
For such $x$ the following holds
\begin{equation}
|(g \circ f)(x) - (g \circ f)(x_0)| = |g(f(x)) - g(f(x_0))| < \epsilon
\end{equation}
which implies that $g \circ f$ is continuous in $x_0$.
\end{proof}
\begin{eg}
Consider the following mappings
\begin{align*}
&f: \realn \longrightarrow \realn, ~~x \longmapsto |x| \\
&g: \realn \longrightarrow \realn \setminus \set{-1}, ~~y \longmapsto \frac{1 - y}{1 + y} \\
&h: \realn \longrightarrow \realn, ~~x \longmapsto \frac{1 - |x|}{1 + |x|}
\end{align*}
It is clear that $h = g \circ f$. Since $f$, $g$ are continuous, $h$ must also be continuous.
\end{eg}
\begin{eg}
The functions $\exp$, $\sin$ and $\cos$ are continuous. We know that
\[
\limes{h}{0} \frac{\exp(k) - 1}{h} = 1
\]
From this follows that
\[
\limes{h}{0} \exp(k) = \exp(0) = 0
\]
Thus, $\exp$ is continuous in $0$. Let $x_0 \in \realn$, then
\begin{align*}
\limes{x}{x_0} \exp(x) &= \limes{h}{0} \exp(x_0 + h) = \limes{h}{0} \exp(x_0)\exp(h) \\
&= \exp(x_0) - \limes{h}{0} \exp(h) = \exp{x_0}
\end{align*}
Now, consider the function $x \mapsto \exp(ix)$. For $x_0 \in \realn$
\begin{align*}
|\underbrace{\exp(i(x_0 + h))}_{\exp(ix_0) \dot \exp(ih)} - \exp(i h_0)| &= \underbrace{|\exp(ix_0)|}_1|\exp(ih) - 1| \\
&\le 1 \cdot \left| \sum_{k = 0}^{\infty} \frac{(ih)^k}{k!} - 1 \right| = \left| \series{k} \frac{(ih)^k}{k!} \right| \\
&\le \series{k} \left| \frac{(ih)^k}{k!} \right| \\
&= \series{k} \frac{|h|^k}{k!} = \sum_{k = 0}^{\infty} \frac{|h|^k}{k!} - 1 = \exp(|h|) - 1
\end{align*}
For $h \rightarrow 0$, the absolute function converges $|h| \rightarrow 0$, and therefore
\[
\lim{h}{0} |\exp(i(x_0 + h)) - \exp(ix)| = 0
\]
due to the squeeze theorem. I.e., $x \mapsto \exp(ix)$ is also continuous. Thus
\begin{align*}
\cos x = \Re(\exp(ix)) && \sin x = \Im(\exp(ix))
\end{align*}
are also continuous due to the concatination of continuous functions.
\end{eg}
\begin{lem}
Let $a, b \in \realn$ with $a < b$, and let
\[
f: [a, b] \longrightarrow \realn
\]
be a continuous function. Furthermore, let $y \in \realn$. Now if the set
\[
\set[f(x) \ge y]{x \in [a, b]}
\]
is non-empty, it has a smallest element.
\end{lem}
\begin{proof}
Let $M$ be non-empty. Set $x_0 = \inf\set{M}$. Then it is to be shown that $x_0 \in M$, or that $f(x_0) \ge y$.
There exists a sequence $\anyseqdef{M}$ such that $x_n \rightarrow x_0$. Because of the continuity of $f$,
\begin{equation}
f(x_0) = f(\limn x_n) = \limn f(x_n) \ge y
\end{equation}
holds, thus $x_0 \in M$.
\end{proof}
\begin{thm}[Extreme value theorem]
Let $a, b \in \realn$ with $a < b$, and let $f: [a, b] \rightarrow \realn$ continuous.
Then the function $f$ attains a maximum, i.e.
\[
\exists x_0 \in [a, b] ~\forall x \in [a, b]: ~~f(x) \le f(x_0)
\]
\end{thm}
\begin{proof}
First we show that $f$ is bounded. Assume $f$ is unbounded above, i.e.
\begin{equation}
\set[f(x) \ge n]{x \in [a, b]} =: M_n, ~~n \in \natn
\end{equation}
According to the last lemma, every $M_n$ has a smallest element $x_n$.
The sequence $(x_n)_{n \in \natn}$ is monotonically increasing ($M_{n+1} \subset M_n$) and bounded above by $b$.
Thus, $x_n$ converges to some $x_0 \in [a, b]$. Now consider the sequence $(f(x_n))_{n \in \natn}$. By definition
\begin{equation}
\limn f(x_n) \ge \limn n = \infty
\end{equation}
And since $f$ is continuous, $\limn f(x_n) = f(x_0)$ must hold. This contradicts the assumption, so $f$ is bounded.
Now set
\begin{equation}
y = \sup\set[{x \in [a, b]}]{f(x)}
\end{equation}
In case $f$ is equal to $y$ everywhere, there is nothing to show. So assume that there are values for which $f \ne y$.
According to the definition of the supremum, the sets
\begin{equation}
\set[f(x) \ge y - \frac{1}{n}]{x \in [a, b]}
\end{equation}
are non-empty for all $n \in \natn$, and thus they have a smallest element $x_n$.
The sequence $(x_n)_{n \in \natn}$ is monotonically increasing and bounded, i.e. it converges to $x_0 \in [a, b]$.
Therefore
\begin{equation}
y \ge f(x_0) = \limn f(x_n) \ge \limn y - \frac{1}{n} = y
\end{equation}
From this follows
\begin{equation}
f(x_0) = y \implies f(x_0) \text{ upper bound of } \set[{x \in [a, b]}]{f(x)}
\end{equation}
\end{proof}
\begin{thm}[Intermediate value theorem]
Let $a, b \in \realn$ with $a < b$, and $f: [a, b] \rightarrow \realn$ a continuous function with $f(a) < f(b)$.
\[
y \in (f(a), f(b)) \implies \exists x_0 \in (a, b): ~~f(x_0) = y
\]
\end{thm}
\begin{proof}
\noproof
\end{proof}
\begin{eg}
$\cos$ has in $[0, 2]$ exactly one root. Consider the definition
\[
\cos x = \sum_{k = 0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}
\]
We know that $\cos 0 = 1$. Furthermore we can show that
\[
-1 = \underbrace{1 - \frac{2^2}{2!}}_{\text{2nd partial sum}} \le \cos(2) \le \underbrace{1 - \frac{2^2}{2!} + \frac{2^4}{4!}}_{\text{3rd partial sum}} < 0
\]
The intermediate value theorem tells us that there exists a root in $[0, 2]$. Now we need to show that $\cos$ is strictly monotonically decreasing on $[0, 2]$.
Choose $z \in [0, 2]$. Then
\[
z \le \sin z \le z - \frac{z^3}{3!}
\]
The addition theorem tells us that
\[
\cos(x) - \cos(y) = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) < 0
\]
for $x, y \in (0, 2]$ and $x > y$. Thus $\cos$ is strictly monotonically decreasing on $[0, 2]$.
\end{eg}
\begin{cor}
Let $I$ be an interval and $f: I \rightarrow \realn$ continuous. Then $f(I)$ is also an interval.
\end{cor}
\begin{proof}
\reader
\end{proof}
\begin{thm}
Let $I$ be an interval, $f: I \rightarrow \realn$ continuous. If $f$ is strictly monotonically increasing, then the inverse function
$\inv{f}: f(I) \rightarrow I$ exists and is continuous.
\end{thm}
\begin{hproof}
$f(I)$ is an interval, and $f$ is injective. This is because if $f(x) = f(\tilde{x})$, then $x = \tilde{x}$ or else $f$ wouldn't be strictly monotonic.
This means
\begin{equation}
\exists g: f(I) \longrightarrow \realn: ~~f(x) = y \iff g(y) = x
\end{equation}
Let $y_0 \in f(I)$ and $\epsilon > 0$. We require that $x_0$ is not a boundary point of $I$. Then choose $0 < \tilde{\epsilon} < \epsilon$ such that
$(x_0 - \tilde{\epsilon}, x_0 + \tilde{epsilon}) \in I$. Choose
\begin{equation}
\delta = \min \set{\underbrace{f(x_0 + \tilde{\epsilon}) - y_0}_{> 0}, \underbrace{y_0 - f(x_0 - \tilde{\epsilon})}_{> 0}} > 0
\end{equation}
If $y \in f(I)$ with $|y - y_0| < \delta$ then
\begin{equation}
f(x_o - \tilde{epsilon}) \le x_0 - \delta < y < y_0 + \delta \le f(x_0 + \tilde{\epsilon})
\end{equation}
From the strict monotony of $g$ we can conclude
\begin{equation}
x_0 - \tilde{epsilon} < g(y) < x_0 + \tilde{\epsilon}
\end{equation}
so
\begin{equation}
|g(y) - g(y_0)| = |g(y) - x_0| < \tilde{\epsilon} < \epsilon
\end{equation}
Thus, $g$ is continuous in $y_0$. Since $y_0 \in f(I)$ was chose arbitrarily, all of $g$ is continuous.
To prove the monotony of $g$, assume $y < \tilde{y}$ and $g(y) \ge g(\tilde{y})$ for $y, \tilde{y} \in f(I)$. From the monotony of $f$ we know that
\begin{equation}
y \ge \tilde{y}
\end{equation}
which is a contradiction, so $g$ is strictly monotonic.
\end{hproof}
\begin{eg}
\begin{enumerate}[(i)]
\item Let $n \in \natn$ and consider
\begin{align*}
f: [0, \infty) &\longrightarrow \realn \\
x &\longmapsto x^n
\end{align*}
$f$ is continuous and strictly monotonically increasing. Thus the inverse function
\[
\sqrt[n]{\cdot}: [0, \infty) \longrightarrow \realn^+
\]
is also continuous.
\item Consider $\exp: \realn \rightarrow \realn$. It's clear that $\exp(\realn) = (0, \infty)$, so the mapping
\[
\ln: (0, \infty) \rightarrow \realn
\]
is continuous and strictly monotonically increasing.
\item Equal arguments can be made for the trigonometric functions.
\end{enumerate}
\end{eg}
\end{document}

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@ -44,6 +44,7 @@
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@ -60,6 +61,7 @@
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