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\noindent{\bf An Example of an Epsilon-Delta Proof}

We say that function $f$ is {\sl continuous\/} at $x$ if
$$(\forall\epsilon >0)(\exists\delta >0)(\forall y\ni |y-x|≤\delta)(|f(y)-f(x)|≤
\epsilon)\,.$$
Diagrammatically, knowing $x$ and $f(x)$,

\bigskip
$$\vcenter{\halign{\hfil$#$\quad&\hfil$#$\hfil\qquad\qquad\qquad\qquad%
&\hfil$#$\hfil\qquad\qquad\qquad\qquad&$#$\hfil\cr
&&&f\cr
\noalign{\bigskip\bigskip\bigskip}
f(x)&-&\bullet\cr
\noalign{\bigskip\bigskip\bigskip}
&&\bullet\cr
\noalign{\smallskip}
&&x\cr}}$$

\bigskip
\noindent
and a vertical range of $\epsilon >0$ either side of $f(x)$

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$$\vcenter{\halign{\hfil$#$\quad&\hfil$#$\hfil\qquad\qquad\qquad\qquad%
&\hfil$#$\hfil\qquad\qquad\qquad\qquad&$#$\hfil\cr
&&&f\cr
\noalign{\bigskip}
f(x)+\epsilon&-\cr
\noalign{\bigskip}
f(x)&-&\bullet\cr
\noalign{\bigskip}
f(x)-\epsilon&-\cr
\noalign{\bigskip}
&&\bullet\cr
\noalign{\smallskip}
&&x\cr}}$$

\bigskip
\noindent
we can find a horizontal range of $\delta$ on either side of~$x$ such that
any point on the graah that is in the horizontal range is also in the
vertical range.

\bigskip
$$\vcenter{\halign{\hfil$#$\quad&\hfil$#$\hfil\qquad\qquad\qquad%
&\hfil$#$\hfil\quad
&\hfil$#$\hfil\quad
&\hfil$#$\hfil\qquad\qquad\qquad&$#$\hfil\cr
f(x)+\epsilon&-\cr
\noalign{\bigskip}
f(x)&-&&\bullet\cr
\noalign{\bigskip}
f(x)-\epsilon&-\cr
\noalign{\bigskip}
&&&\bullet\cr
\noalign{\smallskip}
&&x-\delta&x&x+\delta\cr}}$$

\bigskip
\noindent
The usual way to do this is to express $\delta$ as a function $\delta(x,\epsilon)$;
in other words, show how, given $x$ and~$\epsilon$, to pick~$\delta$.

\bigskip
\noindent{\bf Example:}
We now prove $\sin x$ continuous.

\smallskip\noindent
{\bf Method 1:}
$$\sin y-\sin x=\int↓x↑y\cos \theta\,d\theta\,;$$
since $|\cos\theta|≤1$,
$$-(y-x)=\int↓x↑y(-1)d\theta ≤\int↓x↑y\cos\theta\,d\theta≤\int↓x↑y1\cdot d\theta
=y-x\,,$$
or $|\sin y-\sin x|≤|y-x|$.
(Actually, this assumes $y≥x$, but the contrary case just interchanges
$x$ and~$y$.)

\smallskip\noindent
{\bf Method 2:}


\figbox 2truein:


$$|\sin y-\sin x|=\overline{AC}≤\overline{AB}=|y-x|$$
In either case, by setting $\delta =\epsilon$, $|y-x|<\delta$ implies
$$|\sin y-\sin x|≤|y-x|≤\delta =\epsilon\,.$$

Another example: the product of two continuous functions is continuous.
Let $h(x)=f(x)\cdot g(x)$, both continuous.

\figbox 2truein:

$$h(y)-h(x)=f(x)\bigl(g(y)-g(x)\bigr)+g(x)\bigl(f(y)-f(x)\bigr)+
\bigl(g(y)-g(x)\bigr)(\bigl(f(y)-f(x)\bigr)\,.$$
To keep $h(y)-h(x)≤\epsilon$, we can keep each of the three terms on the
right $≤\epsilon/3$.

To keep $|f(x)\cdot\bigl(g(y)-g(x)\bigr)|≤\epsilon/3$ by keeping
$|x-y|<\delta↓1$, any $\delta↓1$ will do if $f(x)=0$. Otherwise, choose
$\delta↓1$ so that
$$|g(y)-g(x)|≤\left|{\epsilon\over 3f(x)}\right|\,.$$
Similarly choose $\delta↓2$ so that $|f(y)-f(x)|≤\left|{\epsilon\over 
3g(x)}\right|$.
Choose $\delta↓3$ so that $|g(y)-g(x)|≤1$, and $\delta↓4$ so that
$|f(y)-f(x)|≤\epsilon/3$.

Let $\delta$ be the smallest of $\delta↓1$, $\delta↓2$, $\delta↓3$, and
$\delta↓4$; $\delta$~is positive and if $|y-x|<\delta$, $|h(y)-h(x)|
≤{\epsilon\over 3}+{\epsilon\over 3}+1\cdot{\epsilon\over 3}=\epsilon$.

If your calculus course did not teach you how to do such proofs, you were
cheated. I~am serious. Please complain.


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\copyright 1984 Robert W. Floyd

First draft April 22, 1986


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