\documentclass[a4paper,12pt]{article}
\begin{document}

\noindent {\bf Question}

\noindent Prove, using the definition, that each of the following
functions is continuous at all points of ${\bf R}$.
\begin{enumerate}
\item $h_n(x) =x^n$, where $n\in {\bf N}$;
\item $g(x) = c$, where $c\in {\bf R}$;
\item $f$ is a function on ${\bf R}$ which satisfies $|f(x)-f(y)|\leq
c|x-y|$ for all $x$, $y\in {\bf R}$, where $c >0$ is a constant.
\end{enumerate}

\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item To show that $h_n(x)$ is continuous at $a\in {\bf R}$, we need
to show that $\lim_{x\rightarrow a} h_n(x) =h_n(a)$.  Recalling
the definition of limit, this translates to showing that for each
$\varepsilon >0$, there exists $\delta >0$ so that if $| x-a| <
\delta$, then $|h_n(x) -h_n(a)| <\varepsilon$.   Since $h_n(x)
=x^n$, this is the same as showing that for each $\varepsilon >0$,
there exists $\delta >0$ so that if $| x-a| < \delta$, then $| x^n
-a^n| <\varepsilon$.  Let's break the proof into cases.

\medskip
\noindent If $n =1$, then all we need to do to satisfy the
definition is take $\delta =\varepsilon$.  So, we can assume that
$n\ge 2$.  If in addition we have that $a =0$, then by the
definition of limit, we need to show that for each $\varepsilon
>0$, there is $\delta >0$ so that if $|x| <\delta$, then $|x^n| =
|x|^n <\varepsilon$.  So, taking $\delta =\varepsilon^{1/n}$, we
are done in this case as well.

\medskip
\noindent Consider now the case that $n\ge 2$ and $a >0$, and
factor $|x^n -a^n|$ to get $ |x^n -a^n| = | (x-a)(x^{n-1} +
ax^{n-2} +\cdots +a^{n-2}x + a^{n-1})|$.  Recall that we have a
great deal of choice in how we choose $\delta$, so we may restrict
our attention to the interval $|x-a| < \frac{1}{2}a$, so that
$\frac{1}{2} a <x <\frac{3}{2}a$, by requiring that $\delta <
\frac{1}{2}a$ (which makes sense, since $a >0$).  Calculating, we
see that
\begin{eqnarray*}
 |x^n -a^n| & = & | (x-a)(x^{n-1} + ax^{n-2} +\cdots +a^{n-2}x +
  a^{n-1})| \\
 & \le & |x-a| (x^{n-1} + ax^{n-2} +\cdots + a^{n-2} x + a^{n-1})
  \\
 & < & |x-a| \left( \left(\frac{3}{2} a\right)^{n-1} + a
  \left(\frac{3}{2} a\right)^{n-2} +\cdots + a^{n-2} \frac{3}{2} a +
  a^{n-1}\right) \\
 & = & |x-a| a^{n-1}\sum_{k=0}^{n-1} \left( \frac{3}{2}\right)^k \\
 & = & |x-a| a^{n-1}\frac{1 -(3/2)^n}{1-(3/2)} = C |x-a|,
\end{eqnarray*}
where $C =a^{n-1}\frac{1 -(3/2)^n}{1-(3/2)} > 0$ depends on both
$a
>0$ and $n\ge 2$.  So, take $\delta$ to be the smaller of
$\frac{1}{C}\varepsilon$ and $\frac{1}{2}a$.  Then, for $|x-a|
<\delta$, we have that $|x^n -a^n | < C|x-a| \le \varepsilon$ as
desired.  (The first inequality follows from the calculation above
and the fact that $|x-a| < \delta < \frac{1}{2} a$, while the
second inequality follows from $\delta < \frac{1}{C}
\varepsilon$.)

\medskip
\noindent A similar argument, with appropriate placements of
absolute values, holds for $a <0$.  (Note that for a given
$\varepsilon >0$, the choice of $\delta$ depends on $\varepsilon$,
on $a$, and on $n$.)
\item To show that $g(x)$ is continuous at $a\in {\bf R}$, we need
to show that $\lim_{x\rightarrow a} g(x) =g(a)$.  Recalling the
definition of limit, this translates to showing that for each
$\varepsilon >0$, there exists $\delta >0$ so that if $| x-a| <
\delta$, then $|g(x) -g(a)| <\varepsilon$.   Since $g(x) =c$ for
all $x$, this is the same as showing that for each $\varepsilon
>0$, there exists $\delta >0$ so that if $| x-a| < \delta$, then
$| c - c| = 0 <\varepsilon$.  So, regardless of the value of
$\varepsilon$, taking $\delta = 1$ (or whatever your favorite
positive number happens to be today) satisfies the definition.
\item To show that $f(x)$ is continuous at $a\in {\bf R}$, we need
to show that $\lim_{x\rightarrow a} f(x) =f(a)$.  Recalling the
definition of limit, this translates to showing that for each
$\varepsilon >0$, there exists $\delta >0$ so that if $| x-a| <
\delta$, then $|f(x) -f(a)| <\varepsilon$.   Since $|f(x) -f(a)|
\le c|x-a|$, taking $\delta =\frac{1}{c}\varepsilon$ satisfies the
definition.  (If $|x-a| < \delta = \frac{1}{c}\varepsilon$, then
$|f(x) -f(a)| \le c|x-a| < c \frac{1}{c}\varepsilon =\varepsilon$,
as desired.) (Functions that satisfy this condition are often
referred to as {\bf Lipschitz functions}.)
\end{enumerate}

\end{document}