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%MODULE 1: Optimization
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\begin_layout Title
Continuity of Functions
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\begin_layout Subtitle
We look at some harder cases of deciding when a function is continuous.
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Introduction
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{
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 Determining whether a function is continuous on a given domain, requires
 showing that at each point in the domain that the functions is continuous.
 
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\begin_layout Itemize
We know that all polynomials are continuous on 
\begin_inset Formula $\ec\reals$
\end_inset

, and thus they are continuous on any interval 
\begin_inset Formula $\ec[a,b]\subset\reals$
\end_inset

.
\end_layout

\begin_layout Itemize
Similarly, the elementary functions such as 
\begin_inset Formula $\ec\sin(x)$
\end_inset

, 
\begin_inset Formula $\ec\cos(x)$
\end_inset

 and 
\begin_inset Formula $\ec\exp(x)$
\end_inset

 are all continuous on 
\begin_inset Formula $\ec\reals$
\end_inset

.
\end_layout

\begin_layout Itemize
To determine whether a function 
\begin_inset Formula $\ec f$
\end_inset

 is continuous at a point 
\begin_inset Formula $\ec x_{0}$
\end_inset

 in the domain, we have to show that the limit of the function as it approaches
 
\begin_inset Formula $\ec x_{0}$
\end_inset

 from the left equals the limit of the functions as it approaches 
\begin_inset Formula $\ec x_{0}$
\end_inset

 from the right, and these must equal the value of the function at 
\begin_inset Formula $\ec x_{0}$
\end_inset

.
 Formally this is given by, 
\begin_inset Formula 
\[
\ec\lim_{x\rightarrow x_{0}^{-}}f(x)=\lim_{x\rightarrow x_{0}^{+}}f(x)=f(x_{0}).
\]

\end_inset

 Note that this means the function must be defined at 
\begin_inset Formula $\ec x_{0}$
\end_inset

.
 For example the function 
\begin_inset Formula $\ec\ln(x)$
\end_inset

 is not continuous at 
\begin_inset Formula $\ec x=0$
\end_inset

 because the logarithm is not defined there.
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If you need to review this, go to 
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and search for 
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 continuity
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.
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Defining the problem
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 The Problem:
\series default
 Determine on which domain 
\begin_inset Formula 
\[
\ec g(x)=\left\{ \begin{array}{ll}
(x^{2}-4)/(x+2) & \mbox{ if }x<0\\
5 & \mbox{ if }x=0\\
(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4) & \mbox{ if }x>0
\end{array}\right.
\]

\end_inset

 is continuous.
 Can the function 
\begin_inset Formula $\ec g(x)$
\end_inset

 be redefined to have a greater domain of continuity?
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 What does it mean to have a greater, i.e., a larger domain of continuity?
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If 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous on the 
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 set
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\begin_inset Formula $\ec S\subset\reals$
\end_inset

, to have a greater domain of discontinuity means that the 
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 set
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 of points on which 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous can be extended to a larger set 
\begin_inset Formula $\ec\tilde{S}$
\end_inset

 with 
\begin_inset Formula $\ec S\subset\tilde{S}$
\end_inset

, such that there is a function, we'll call it 
\begin_inset Formula $\ec\tilde{g}(x)$
\end_inset

, that is continuous on on 
\begin_inset Formula $\ec\tilde{S}$
\end_inset

, 
\shape italic
and
\shape default
 
\begin_inset Formula $\ec\tilde{g}(x)=g(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x\in S$
\end_inset

.
 
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\begin_layout Itemize
First, if 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous on 
\begin_inset Formula $\ec\reals$
\end_inset

, then 
\begin_inset Formula $\ec S=\reals$
\end_inset

, and it is impossible to increase the size of this set.
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Are there any other points of discontinuity?
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%1
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Examining the function
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\begin_layout Itemize
The possible points of discontinuity are when 
\begin_inset Formula $\ec x=0$
\end_inset

, (the break point in the definition of 
\begin_inset Formula $\ec g(x)$
\end_inset

), and at the points at which the denominators are zero, i.e., where 
\begin_inset Formula $\ec(x+2)=0$
\end_inset

, and when 
\begin_inset Formula $\ec(x^{2}+4x+4)=0$
\end_inset

.
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\begin_layout Itemize
The first of equations can be solved immediately to give 
\begin_inset Formula $\ec x=-2$
\end_inset

, while the second requires only a bit more work in 
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 factoring
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 the quadratic polynomial into the terms 
\begin_inset Formula $\ec(x+2)(x+2)$
\end_inset

.
 Thus 
\begin_inset Formula $\ec x=-2$
\end_inset

 and 
\begin_inset Formula $\ec x=2$
\end_inset

 are possible points where 
\begin_inset Formula $\ec g(x)$
\end_inset

 may fail to be defined.
 Thus, 
\begin_inset Formula 
\begin{align*}
 & \textrm{Possible point}\\
 & \textrm{ of discontinuity}\\
\ec g(x)=\left\{ \begin{array}{ll}
(x^{2}-4)/(x+2), & \mbox{ if }x<0\\
5, & \mbox{ if }x=0\\
\frac{x^{3}+2x^{2}-2x-4}{x^{2}+4x+4}, & \mbox{ if }x>0
\end{array}\right. & \begin{array}{l}
\implies\mbox{{\rm at }}x=-2\\
\implies\mbox{{\rm at }}x=0\\
\implies\mbox{{\rm at }}x=-2
\end{array}
\end{align*}

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We should graph each 
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 branch
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 of the function 
\begin_inset Formula $\ec g(x)$
\end_inset

 so that we have a sense of what his going on.
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Plotting of 
\begin_inset Formula $g(x)$
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\begin_layout Itemize
We graph the the branches of the function, 
\begin_inset Formula $\ec f(x)=g(x)$
\end_inset

,
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\begin_deeper
\begin_layout Standard
\begin_inset Formula $\ec g_{1}(x)=(x^{2}-4)/(x+2)$
\end_inset

, 
\begin_inset Newline newline
\end_inset

 
\begin_inset Formula $\ec g_{2}(x)=5$
\end_inset

, and
\begin_inset Newline newline
\end_inset

 
\begin_inset Formula $\ec g_{3}(x)=(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)$
\end_inset


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using 
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to input the value 
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 (x**3 + 2*x**2 - 2*x -4)/(x**2 + 4*x + 4), 5, (x*x-4)/(x+2)
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 where the 
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 independent
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 variable is 
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 x
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.
 
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 You will need to examine the figure you obtain closely.
 Try to understand what each line represents, and how they relate to the
 three branches of the function 
\begin_inset Formula $\ec g(x)$
\end_inset

.
 Try to identify the possible singularities, and points of discontinuity
 of the function 
\begin_inset Formula $\ec g(x)$
\end_inset

.
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The graph of 
\begin_inset Formula $g(x)$
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\begin_inset Caption

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 Note that function 
\begin_inset Formula $\ec g(x)=g_{1}(x)$
\end_inset

 for all 
\begin_inset Formula $\ec x<0$
\end_inset

 appears to be the equation of a line, i.e., a function of the form 
\begin_inset Formula $y=mx+b$
\end_inset

, and that the function 
\begin_inset Formula $\ec g_{3}(x)$
\end_inset

 has a singularity at 
\begin_inset Formula $\ec x=-2$
\end_inset

.
 The value of 
\begin_inset Formula $\ec g(x)$
\end_inset

 at 
\begin_inset Formula $\ec x=0$
\end_inset

, i.e., 
\begin_inset Formula $\ec g(0)$
\end_inset

, is a single point that is not connected to the values of 
\begin_inset Formula $\ec g(x)$
\end_inset

 to the left or to the right of this point.
 
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Examining 
\begin_inset Formula $g(x)$
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 on 
\begin_inset Formula $(-\infty,0)$
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\begin_layout Itemize
If 
\begin_inset Formula $\ec x<0$
\end_inset

 then: 
\begin_inset Formula 
\[
\ec g(x)=\frac{(x^{2}-4)}{(x+2)}=\frac{(x-2)(x+2)}{(x+2)}=g_{1}(x).
\]

\end_inset

 
\end_layout

\begin_layout Itemize
This function is not continuous since the function is not defined at 
\begin_inset Formula $\ec x=-2$
\end_inset

.
 However, since 
\begin_inset Formula 
\begin{align*}
\ec\lim_{x\rightarrow-2^{-}}g(x) & =\frac{(x-2)(x+2)}{(x+2)}=\lim_{x\rightarrow-2^{-}}(x-2)=-4\\
\ec\lim_{x\rightarrow-2^{+}}g(x) & =\frac{(x-2)(x+2)}{(x+2)}=\lim_{x\rightarrow-2^{+}}(x-2)=-4
\end{align*}

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 the 
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 limit from the left
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 equals the 
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{
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 limit from the right
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}
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.
 This is typical of a 
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 removable discontinuity
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\end_inset

.
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\begin_layout Itemize
Thus, if we define 
\begin_inset Formula $\ec\tilde{g}(x)=(x-2)$
\end_inset

, then 
\begin_inset Formula $\ec\tilde{g}(x)=g(x)$
\end_inset

 on 
\begin_inset Formula $\ec(-\infty,0)$
\end_inset

 except at 
\begin_inset Formula $\ec x=-2$
\end_inset

, and 
\begin_inset Formula $\ec\tilde{g}$
\end_inset

 is continuous on 
\begin_inset Formula $\ec(-\infty,0)$
\end_inset

.
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Note that 
\begin_inset Formula $\ec g_{1}(x)=x-2$
\end_inset

 is the equation of a line, as we observed in the graph on 
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page:cont-graph
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The details of taking the limit
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\begin_layout Itemize
The evaluation the limit from below 
\begin_inset Formula ${\displaystyle \ec\lim_{x\rightarrow-2^{-}}\frac{(x-2)(x+2)}{(x+2)}=-4}$
\end_inset

, in 
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eq:blim
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 and similarly the limit in 
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eq:alim
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 seems obvious.
 We simply cancel the terms 
\begin_inset Formula ${\displaystyle \ec\frac{(x+2)}{(x+2)}=1}$
\end_inset

.
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\begin_layout Itemize
There is a technical issue here, i.e., at the limit point, 
\begin_inset Formula $\ec x=-2$
\end_inset

, we have that 
\begin_inset Formula ${\displaystyle \ec\frac{(x+2)}{(x+2)}=\frac{0}{0}}$
\end_inset

, and we know that we cannot evaluate this indeterminate form.
\end_layout

\begin_layout Itemize
We can argue that for each 
\begin_inset Formula $\ec x_{0}$
\end_inset

 in any 
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 neighborhood
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 of 
\begin_inset Formula $\ec x=-2$
\end_inset

 we have that 
\begin_inset Formula $\ec(x_{0}+2)/(x_{0}+2)=1$
\end_inset

 and thus the limit must be 
\begin_inset Formula $\ec1$
\end_inset

, and thus in any neighborhood of 
\begin_inset Formula $\ec x=-2$
\end_inset

 the value of 
\begin_inset Formula $\ec{\displaystyle \frac{(x-2)(x+2)}{(x+2)}=1\cdot(x-2)-4}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

As long as we are not at 
\begin_inset Formula $\ec x=-2$
\end_inset

, we can evaluate the expression 
\begin_inset Formula $\ec(x+2)/(x+2)=1$
\end_inset

, and this is all that matters in taking the limit.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Examining 
\begin_inset Formula $g(x)$
\end_inset

 on 
\begin_inset Formula $(0,\infty)$
\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec x>0$
\end_inset

 then 
\begin_inset Formula 
\[
\ec g(x)=(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)=\frac{(x^{2}-2)(x+2)}{(x+2)^{2}}=\frac{x^{2}-2}{x+2}
\]

\end_inset

 and 
\begin_inset Formula $\ec g(x)$
\end_inset

 has no 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 singularities
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 on the domain 
\begin_inset Formula $\ec(0,\infty)$
\end_inset

 since the point 
\begin_inset Formula $\ec x=-2$
\end_inset

, i.e., where the denominator vanishes, is outside the domain of definition
 of this branch of the function.
\end_layout

\begin_layout Itemize
Note that 
\begin_inset Formula 
\[
\ec(x^{3}+2x^{2}-2x-4)/(x^{2}+4x+4)=\frac{x^{2}-2}{x+2}
\]

\end_inset

 represent the same functions on 
\begin_inset Formula $\ec(0,\infty)$
\end_inset

, i.e., at each point 
\begin_inset Formula $\ec x\in(0,\infty)$
\end_inset

 both of these are algebraically equivalent yielding the same value.
\end_layout

\begin_layout Itemize
Thus 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous on 
\begin_inset Formula $\ec(0,\infty)$
\end_inset

 since 
\begin_inset Formula $\ec\frac{(x^{2}-2)(x+2)}{(x+2)^{2}}$
\end_inset

 is a 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 rational function
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 with no singularities in it its domain.
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

%1
\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Examining 
\begin_inset Formula $g(x)$
\end_inset

 in a neighborhood of 
\begin_inset Formula $x=0$
\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec x=0$
\end_inset

, then consider 
\begin_inset Formula 
\[
\ec\lim_{x\rightarrow0^{-}}g(x)=\lim_{x\rightarrow0^{-}}x-2=-2,
\]

\end_inset

 and 
\begin_inset Formula 
\[
\ec\lim_{x\rightarrow0^{+}}g(x)=\lim_{x\rightarrow0^{+}}\frac{x^{2}-2}{x+2}=-1.
\]

\end_inset


\end_layout

\begin_layout Itemize
Thus 
\begin_inset Formula $\ec\lim_{x\rightarrow0^{-}}g(x)\ne\lim_{x\rightarrow0^{+}}g(x)$
\end_inset

 and 
\begin_inset Formula $\ec g(x)$
\end_inset

 is not continuous at 
\begin_inset Formula $\ec x=0$
\end_inset

.
 This is not a 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 removable discontinuity
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 since we cannot redefine 
\begin_inset Formula $\ec g(0)$
\end_inset

 in a manner which makes the limit from the left equal the limit from the
 right, and which also makes this limit equal to some value of 
\begin_inset Formula $\ec g$
\end_inset

 that we can assign at 
\begin_inset Formula $\ec x=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

%2
\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Itemize
The point 
\begin_inset Formula $\ec x=0$
\end_inset

 is an 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 essential
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 singularity of 
\begin_inset Formula $\ec g(x)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

From everything we have done so far, the conclusion is that 
\begin_inset Formula $\ec g(x)$
\end_inset

 as defined can be made continuous on the domain 
\begin_inset Formula $\ec(-\infty,0)\bigcup(0,\infty)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
vfil
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hfil
\end_layout

\end_inset

 Let's pull this all together.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

%2
\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Redefining 
\begin_inset Formula $g(x)$
\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout OverlayArea
\begin_inset ERT
status open

\begin_layout Plain Layout

{
\backslash
textwidth}{6in}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
We have ascertained that 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous on 
\begin_inset Formula $\ec\reals$
\end_inset

, except at 
\begin_inset Formula $\ec x=-2$
\end_inset

, and 
\begin_inset Formula $\ec x=0$
\end_inset

.
\end_layout

\begin_layout Itemize
We showed that if 
\begin_inset Formula $\ec g(x)$
\end_inset

 is redefined to be 
\begin_inset Formula $\ec\tilde{g}(x)$
\end_inset

 on 
\begin_inset Formula $\ec(-\infty,0)$
\end_inset

, then 
\begin_inset Formula $\ec g(x)$
\end_inset

 is continuous on 
\begin_inset Formula $\ec(-\infty,0)$
\end_inset

, but that at 
\begin_inset Formula $\ec x=0$
\end_inset

, there was no way to redefine 
\begin_inset Formula $\ec g$
\end_inset

 so as to make it continuous at that point.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
If we define 
\begin_inset Formula 
\[
\ec\tilde{g}(x)=\left\{ \begin{array}{ll}
\ec x-2 & \ec\mbox{ if }x<0\\
\ec-2 & \ec\mbox{ if }x=0\\
\ec(x^{2}-2)/(x+2) & \ec\mbox{ if }x>0
\end{array}\right.
\]

\end_inset

 then 
\begin_inset Formula $\ec\tilde{g}(x)=g(x)$
\end_inset

 except at 
\begin_inset Formula $\ec x=-2$
\end_inset

 and 
\begin_inset Formula $\ec x=0$
\end_inset

 and 
\begin_inset Formula $\ec\tilde{g}$
\end_inset

 is continuous on 
\begin_inset Formula $\ec(-\infty,0]\bigcup(0,\infty)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
If we define 
\begin_inset Formula 
\[
\ec\overline{g}(x)=\left\{ \begin{array}{ll}
\ec x-2 & \ec\mbox{ if }x<0\\
\ec-1 & \ec\mbox{ if }x=0\\
\ec(x^{2}-2)/(x+2) & \ec\mbox{ if }x>0
\end{array}\right.
\]

\end_inset

 then 
\begin_inset Formula $\ec\overline{g}(x)=g(x)$
\end_inset

 except at 
\begin_inset Formula $\ec x=-2$
\end_inset

 and 
\begin_inset Formula $\ec x=0$
\end_inset

 and 
\begin_inset Formula $\ec\overline{g}$
\end_inset

 is continuous on 
\begin_inset Formula $\ec(-\infty,0)\bigcup[0,\infty)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<4>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
There is no way to redefine 
\begin_inset Formula $\ec g$
\end_inset

 to be continuous on 
\begin_inset Formula $\ec(-\infty,\infty)$
\end_inset

, and thus 
\begin_inset Formula $\ec\tilde{g}$
\end_inset

 or 
\begin_inset Formula $\ec\overline{g}$
\end_inset

 are the only 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 continuous extensions
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 of the function 
\begin_inset Formula $\ec g$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Conclusion
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Itemize
Some functions can be 
\emph on
extended
\emph default
 in such a way that some, or even all, of their discontinuities are 
\begin_inset Quotes eld
\end_inset

removed
\begin_inset Quotes erd
\end_inset

.
 
\end_layout

\begin_layout Itemize
To determine which discontinuities are 
\emph on
removable
\emph default
 and which are 
\emph on
essential
\emph default
, we must examine the limit from each side: from the left or from the right.
 
\end_layout

\begin_layout EndFrame

\end_layout

\end_body
\end_document
