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Fix descriptions in 1.5 and 1.6
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sean-fitzpatrick committed Oct 7, 2024
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76 changes: 42 additions & 34 deletions ptx/sec_limit_continuity.ptx
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<image width="47%">
<description>
<p>
Shows a graph with domain <m>0</m> to <m>3</m>. For <m>0 \geq x \lt 1</m>
the graph has a downward curve to it, for <m>1 &gt; x \leq 2</m> the graph
is a straight line, parallel with the <m>x</m> axis, and for <m>2 \geq x \leq 3</m>.
The graph is undefined at <m>x = 1</m>.
The graph of a piecewise-defined function is shown, for <m>x</m> from <m>0</m> to <m>3</m>.
For <m>0 \leq x \lt 1</m>
the graph looks like a parabola opening downward.
This part of the graph approaches, but does not reach, the point <m>(1,1)</m>.
There is a hollow dot at <m>(1,1)</m>, indicating that <m>f(1)</m> is undefined.
For <m>1 \lt x \leq 2</m> the graph
is a horizontal line segment, with <m>y=1</m>.
For <m>2 \leq x \leq 3</m> the graph again has the appearance of a downward-facing parabola
that begins at <m>(2,1)</m> and ends at <m>(3,1)</m>.
</p>
</description>
<shortdescription>
Example of a discontinuous graph, where the discontinuity is represented by a hollow dot.
Graph of a function with a discontinuity when x=1. Although the limit at 1 exists, f(1) is undefined.
</shortdescription>
<latex-image label="img_continuous1">
\begin{tikzpicture}[declare function = {func(\x) = (\x &gt;= 0)*(\x &lt;= 1)*(-(\x-1/4)^2+1/16+1.5) + (\x &gt; 1)*(\x &lt;= 2) + (\x &gt; 2)*(\x &lt;= 3)*((2-\x)*(\x-3)+1);}]
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<image width="47%">
<description>
<p>
Shows a graph with domain <m>-2</m> to <m>3</m>. There are five
straight lines, each parallel with the <m>x</m> axis. Each of the lines is
one unit in length and undefined on its right side, but defined on
their left. Line one is defined by the points <m>(-2, -2), (-1, -2)</m>,
line two <m>(-1, -1), (0, -1)</m>, line <m>3</m> <m>(0, 0), (1, 0)</m>, line <m>4</m>
<m>(1, 1), (2, 1)</m>, and line <m>5</m> <m>(2, 2), (3, 2)</m>.
Shows the graph of the greatest integer function, for <m>x</m> from <m>-2</m> to <m>3</m>. There are five
horizontal line segments in a <q>staircase</q> configuration, ascending from left to right. Each segment is
one unit in length and includes its left endpoint, but the right endpoint of each segment is not included.
The first segment is from <m>(-2, -2)</m> to <m>(-1, -2)</m>,
the second from <m>(-1, -1)</m> to <m>(0, -1)</m>, the third from <m>(0, 0)</m> to <m>(1, 0)</m>,
the fourth from <m>(1, 1)</m> to <m>(2, 1)</m>, and the fifth from <m>(2, 2)</m> to <m>(3, 2)</m>.
</p>
</description>
<shortdescription>
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<image>
<description>
<p>
Shows the graph of a function on the domain <m>0</m> to <m>4</m>.
The graph is has a downward curve
and for <m>x = 2</m> the function is defined by the point <m>(2, 1)</m>.
The point <m>(2, 1)</m> shows a removable discontinuity because the graph
is undefined at <m>x = 2</m>, but the point shows that the function is defined
at <m>x = 2</m>.
A portion of the graph of a function is shown, for <m>x</m> from <m>0</m> to <m>4</m>.
The graph has the shape of a parabola opening downward,
but at <m>x=2</m> there is a hole in the graph,
and instead the point <m>(2,1)</m> (which is not on the graph) is plotted.
The graph of this function illustrates a removable discontinuity because
<m>\lim_{x\to 2}f(x)</m> exists, but does not equal <m>f(2)</m>.
</p>
</description>
<shortdescription>
Graph showing a removable discontinuity by having the point (2, 1)
where f(2) is undefined without the point.
Graph showing a removable discontinuity: a hole in the graph when x=2 shows that the limit and function values disagree.
</shortdescription>
<latex-image label="img_discontinuity_rem">
\begin{tikzpicture}
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<image>
<description>
<p>
Shows the graph of a function on the domain <m>0</m> to <m>4</m>.
When approching <m>x = 2</m> from the left hand side <m>f(x)</m> is
undefined, but when coming from the right hand side <m>f(x) = 1</m>. For the
interval <m>0 \lt x \leq 2</m> the graph is curved downward and for the
The graph of a function is shown for <m>x</m> from <m>0</m> to <m>4</m>.
As <m>x</m> approaches <m>2</m> from the left,
the graph of <m>f</m> approaches a point that is not part of the graph,
as indicated by a hollow dot.
As <m>x</m> approaches <m>2</m> from the right,
the graph of <m>f</m> approaches a point that is part of the graph,
as indicated by a solid dot.
The point marked by the solid dot lies below the point marked by the hollow dot,
illustrating that the left and right hand limits are different as <m>x\to 2</m>.
</p>

<p>
On the interval <m>0 \lt x \leq 2</m> the graph is curved downward and on the
interval <m>2 \leq x \leq 4</m> the graph is a straight line with a
positive slope. Because <m>f(x)</m> is undefined at <m>x = 2</m>
when coming from the left, but is defined coming from the right,
there is a jump discontinuity.
positive slope.
</p>
</description>
<shortdescription>
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<image>
<description>
<p>
Shows the graph of a function on the domain <m>0</m> to <m>4</m>.
The graph of a function is shown for <m>x</m> from <m>0</m> to <m>4</m>.
There is a
vertical dotted line at <m>x = 2</m> illustrating an asymptote.
As <m>x</m> approaches <m>2</m> from both sides the value of <m>f(x)</m>
approaches infinity. The graph also has an asymptote at <m>y = 0</m>.
On both sides of the dotted vertical line the graph has an upward
curve with an increasing slope at <m>x</m> gets closer to <m>2</m>.
The asymptote at <m>x = 2</m> causes there to be an infinite
discontinuity.
vertical dotted line at <m>x = 2</m> illustrating a vertical asymptote.
As <m>x</m> approaches <m>2</m> from either side,
the graph of <m>f</m> extends upward along the asymptote,
indicating that the value of <m>f(x)</m> is increasing without bound.
</p>
</description>
<shortdescription>
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69 changes: 42 additions & 27 deletions ptx/sec_limit_infty.ptx
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<p>
Graph of <m>f(x)=1/x^2</m> for <m>x</m> between <m>-1</m> and <m>1</m>. There is a vertical
asymptote at <m>x = 0</m> and a horizontal asymptote at <m>y = 0</m>.
For <m>x</m> values near <m>\infty</m> and <m>-\infty</m>, <m>y</m>
approaches <m>0</m>. For <m>x</m> values near <m>0</m>, <m>y</m> approaches <m>\infty</m>.
For <m>x</m> values near the left and right edges of the image,
the <m>y</m> value is close to <m>0</m>.
For <m>x</m> values near <m>0</m>, the graph extends to the top of the image (and presumably beyond),
suggesting that <m>y</m> approaches <m>\infty</m>.
</p>
</description>
<shortdescription>
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Graph of <m>f(x)=\frac{1}{(x-1)^2}</m> for <m>x</m> between <m>0</m> and <m>1</m>.
There is a vertical
asymptote at <m>x = 1</m> and a horizontal asymptote at <m>y = 0</m>.
As <m>x</m> gets near <m>1</m> from both sides of the vertical asymptote
<m>y</m> approaches <m>\infty</m>. As <m>x</m> gets near <m>\infty</m>
and <m>-\infty</m> <m>y</m> approaches <m>0</m>.
As <m>x</m> gets near <m>1</m> from either side of the vertical asymptote,
<m>y</m> approaches <m>\infty</m>.
For <m>x</m> values near the left and right edges of the image,
the value of <m>y</m> approaches <m>0</m>.
</p>
</description>
<shortdescription>
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approaches <m>-\infty</m> and from the right, <m>y</m> approaches
<m>\infty</m>. As <m>y</m> approaches <m>0</m> from the bottom, <m>x</m>
approaches <m>-\infty</m> and from the top, <m>x</m> approaches
<m>\infty</m>. The lines of the equation are in quadrants one and
three of the graph.
<m>\infty</m>. The graph conists of two parts; one in quadrant one and the other in
quadrant three.
</p>
</description>
<shortdescription>
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<image width="47%">
<description>
<p>
The graph
is a single straight line with a positive slope. At <m>x = 1</m>
there is a hollow point indicating a discontinuity, the exact
position of the discontinuity is <m>(1, 2)</m>.
The graph is a single straight line with a positive slope.
At <m>x = 1</m> there is a hollow dot indicating a removable discontinuity.
The exact position of the discontinuity is <m>(1, 2)</m>.
</p>
</description>
<shortdescription>
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<image>
<description>
<p>
Graph of <m>f(x)=\frac{x^2}{x^2+4}</m> on <m>[-20,20]</m>. There
is a horizontal asymptote at <m>y = 1</m>. As <m>x</m> approaches
<m>-\infty</m> and <m>\infty</m>, <m>f(x)</m> gets near <m>1</m>,
but never equals <m>1</m>. The graph drops to the point <m>(0,0)</m>
Graph of <m>f(x)=\frac{x^2}{x^2+4}</m> showing <m>x</m> values from <m>-20</m> to <m>20</m>.
There is a horizontal asymptote at <m>y = 1</m>.
As <m>x</m> approaches <m>-\infty</m> and <m>\infty</m>, <m>f(x)</m> gets near <m>1</m>,
but never equals <m>1</m>. The graph lies between <m>y=0</m> and <m>y=1</m>,
and drops to the point <m>(0,0)</m>
as <m>x</m> approaches <m>0</m> from either direction.
</p>
</description>
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<image>
<description>
<p>
As
<m>x</m> approaches <m>-\infty</m> and <m>\infty, f(x)</m> gets near
<m>0</m>. The graph dips to a minimum value just to the left of the <m>y</m> axis,
As <m>x</m> approaches <m>-\infty</m> and <m>\infty</m>, <m>f(x)</m> gets near <m>0</m>.
The graph dips to a minimum value just to the left of the <m>y</m> axis,
then crosses the <m>x</m> axis at <m>(0,0)</m>, rising to a maximum value just to the right of the <m>x</m> axis,
before falling again toward the horizontal asymptote <m>y=0</m>.
</p>
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<description>
<p>
The graph of <m>f(x)=\frac{x}{\sqrt{x^2+1}}</m>, which has two horizontal asymptotes,
one at <m>y = -1</m> and the other at <m>y = 1</m>.
one at <m>y = -1</m> (representing the limit as <m>x\to -\infty</m>)
and the other at <m>y = 1</m> (representing the limit as <m>x\to\infty</m>).
</p>
</description>
<shortdescription>
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<image>
<description>
<p>
There
is a horizontal asymptote at <m>y = -\frac{1}{3}</m>, helping
illustrate that the limit of <m>f(x) = \frac{1}{3}</m>,
which is the coefficient of the numerator with the highest
power of <m>x</m> in <m>\dfrac{x^2+2x-1}{1-x-3x^2}</m>
divided by the coefficient of the dennominator with the
highest power of <m>x</m> in <m>\dfrac{x^2+2x-1}{1-x-3x^2}</m>.
The graph of <m>f(x)=\frac{x^2+2x-1}{1-x-3x^2}</m> is shown for <m>x\gt 0</m>.
The graph has a horizontal asymptote at <m>y = -\frac{1}{3}</m>,
which it approaches from below.
The graph illustrates that the limit of <m>f(x)</m> as <m>x\to\infty</m> is <m>-\frac{1}{3}</m>.
The coefficient of the term in the numerator of <m>f(x)</m> with the highest
power of <m>x</m> is <m>1</m>,
while the coefficient of the term in the dennominator with the
highest power of <m>x</m> is <m>-3</m>.
The ratio of these two coefficients gives the limit as <m>x\to\pm \infty</m>
when the highest power of <m>x</m> is the same in both the numerator and the denominator.
</p>
</description>
<shortdescription>
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<!-- START figures/fig_hzasy7.tex -->
<image>
<description>
<p>
The graph of <m>f(x)=\frac{x^2-1}{3-x}</m> is shown for <m>x\gt 3</m>.
Near <m>x=3</m> the graph appears to be heading down a vertical asymptote,
suggesting that <m>\lim_{x\to 3^+}f(x)=-\infty</m>.
The graph then rises to a peak, before beginning to descend again.
Beyond <m>x=10</m>, the graph appears almost straight,
and continues downward at a slope close to <m>-1</m>,
showing that there is no horizontal asymptote in this case.
</p>
<p>
The graph shows that the limit of <m>f(x)</m> will be determined
by dominant terms from the numerator and denominator, which are
<m>x^2</m> and <m>-x</m>. Since <m>\frac{x^2}{-x} = -x</m> for large values
of <m>x</m>, the graph of <m>f(x)</m> behaves approximately the same as that of <m>y=x</m>.
of <m>x</m>, the graph of <m>f(x)</m> behaves approximately the same as that of <m>y=-x</m>.
</p>
</description>
<shortdescription>
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