diff --git a/ptx/docinfo.ptx b/ptx/docinfo.ptx index 889d4d996..486cfc0a5 100644 --- a/ptx/docinfo.ptx +++ b/ptx/docinfo.ptx @@ -163,21 +163,21 @@ \pgfplotsset{hollowdot/.style={color=firstcolor,fill=white,only marks,mark=*}} \pgfplotsset{soliddot/.style={color=firstcolor,fill=firstcolor,only marks,mark=*}} - \pgfplotsset{open/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle[open]}}} - \pgfplotsset{openclosed/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle}}} - \pgfplotsset{closed/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle}}} - \pgfplotsset{closedopen/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle[open]}}} - \pgfplotsset{infiniteopen/.style={firstcurvestyle,shorten >=-2.4pt,{Kite}-{Circle[open]}}} - \pgfplotsset{openinfinite/.style={firstcurvestyle,shorten <=-2.4pt,{Circle[open]}-{Kite}}} - \pgfplotsset{infiniteclosed/.style={firstcurvestyle,shorten >=-2.4pt,{Kite}-{Circle}}} - \pgfplotsset{closedinfinite/.style={firstcurvestyle,shorten <=-2.4pt,{Circle}-{Kite}}} - \pgfplotsset{infinite/.style={firstcurvestyle,{Kite}-{Kite}}} - \pgfplotsset{infiniteleft/.style={firstcurvestyle,{Kite}-}} - \pgfplotsset{infiniteright/.style={firstcurvestyle,-{Kite}}} - \pgfplotsset{openleft/.style={firstcurvestyle,shorten <=-2.4pt,{Circle[open]}-}} - \pgfplotsset{openright/.style={firstcurvestyle,shorten >=-2.4pt,-{Circle[open]}}} - \pgfplotsset{closedleft/.style={firstcurvestyle,shorten <=-2.4pt,{Circle}-}} - \pgfplotsset{closedright/.style={firstcurvestyle,shorten >=-2.4pt,-{Circle}}} + \pgfplotsset{open/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle[open]}}} + \pgfplotsset{openclosed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle}}} + \pgfplotsset{closed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle}}} + \pgfplotsset{closedopen/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle[open]}}} + \pgfplotsset{infiniteopen/.style={shorten >=-2.4pt,{Kite}-{Circle[open]}}} + \pgfplotsset{openinfinite/.style={shorten <=-2.4pt,{Circle[open]}-{Kite}}} + \pgfplotsset{infiniteclosed/.style={shorten >=-2.4pt,{Kite}-{Circle}}} + \pgfplotsset{closedinfinite/.style={shorten <=-2.4pt,{Circle}-{Kite}}} + \pgfplotsset{infinite/.style={{Kite}-{Kite}}} + \pgfplotsset{infiniteleft/.style={{Kite}-}} + \pgfplotsset{infiniteright/.style={-{Kite}}} + \pgfplotsset{openleft/.style={shorten <=-2.4pt,{Circle[open]}-}} + \pgfplotsset{openright/.style={shorten >=-2.4pt,-{Circle[open]}}} + \pgfplotsset{closedleft/.style={shorten <=-2.4pt,{Circle}-}} + \pgfplotsset{closedright/.style={shorten >=-2.4pt,-{Circle}}} \pgfplotsset{every axis/.append style = { cycle list name = curvestylelist, @@ -282,21 +282,21 @@ \pgfplotsset{hollowdot/.style={color=firstcolor,fill=white,only marks,mark size=2.4pt,mark=*}} \pgfplotsset{soliddot/.style={color=firstcolor,fill=firstcolor,only marks,mark size=2.4pt,mark=*}} - \pgfplotsset{open/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle[open]}}} - \pgfplotsset{openclosed/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle}}} - \pgfplotsset{closed/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle}}} - \pgfplotsset{closedopen/.style={firstcurvestyle,shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle[open]}}} - \pgfplotsset{infiniteopen/.style={firstcurvestyle,shorten >=-2.4pt,{Kite}-{Circle[open]}}} - \pgfplotsset{openinfinite/.style={firstcurvestyle,shorten <=-2.4pt,{Circle[open]}-{Kite}}} - \pgfplotsset{infiniteclosed/.style={firstcurvestyle,shorten >=-2.4pt,{Kite}-{Circle}}} - \pgfplotsset{closedinfinite/.style={firstcurvestyle,shorten <=-2.4pt,{Circle}-{Kite}}} - \pgfplotsset{infinite/.style={firstcurvestyle,{Kite}-{Kite}}} - \pgfplotsset{infiniteleft/.style={firstcurvestyle,{Kite}-}} - \pgfplotsset{infiniteright/.style={firstcurvestyle,-{Kite}}} - \pgfplotsset{openleft/.style={firstcurvestyle,shorten <=-2.4pt,{Circle[open]}-}} - \pgfplotsset{openright/.style={firstcurvestyle,shorten >=-2.4pt,-{Circle[open]}}} - \pgfplotsset{closedleft/.style={firstcurvestyle,shorten <=-2.4pt,{Circle}-}} - \pgfplotsset{closedright/.style={firstcurvestyle,shorten >=-2.4pt,-{Circle}}} + \pgfplotsset{open/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle[open]}}} + \pgfplotsset{openclosed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle[open]}-{Circle}}} + \pgfplotsset{closed/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle}}} + \pgfplotsset{closedopen/.style={shorten <=-2.4pt,shorten >=-2.4pt,{Circle}-{Circle[open]}}} + \pgfplotsset{infiniteopen/.style={shorten >=-2.4pt,{Kite}-{Circle[open]}}} + \pgfplotsset{openinfinite/.style={shorten <=-2.4pt,{Circle[open]}-{Kite}}} + \pgfplotsset{infiniteclosed/.style={shorten >=-2.4pt,{Kite}-{Circle}}} + \pgfplotsset{closedinfinite/.style={shorten <=-2.4pt,{Circle}-{Kite}}} + \pgfplotsset{infinite/.style={{Kite}-{Kite}}} + \pgfplotsset{infiniteleft/.style={{Kite}-}} + \pgfplotsset{infiniteright/.style={-{Kite}}} + \pgfplotsset{openleft/.style={shorten <=-2.4pt,{Circle[open]}-}} + \pgfplotsset{openright/.style={shorten >=-2.4pt,-{Circle[open]}}} + \pgfplotsset{closedleft/.style={shorten <=-2.4pt,{Circle}-}} + \pgfplotsset{closedright/.style={shorten >=-2.4pt,-{Circle}}} \pgfplotsset{every axis/.append style = { cycle list name = curvestylelist, diff --git a/ptx/sec_alt_series.ptx b/ptx/sec_alt_series.ptx index ea412b870..e28cf72c5 100644 --- a/ptx/sec_alt_series.ptx +++ b/ptx/sec_alt_series.ptx @@ -211,13 +211,13 @@ x post scale=2, ] - \addplot[infiniteright,domain=0:1] ({x},{2}) node [pos=1, right] {$a_1$}; - \addplot[infiniteleft,domain=0.333:1] ({x},{1.75}) node [pos=0, left] {$-a_2$}; - \addplot[infiniteright,domain=0.333:.833] ({x},{1.5}) node [pos=1, right] {$a_3$}; - \addplot[infiniteleft,domain=0.433:.8333] ({x},{1.25}) node [pos=0, left] {$-a_4$}; - \addplot[infiniteright,domain=0.4333:.76667] ({x},{1}) node [pos=1, right] {$a_5$}; - \addplot[infiniteleft,domain=0.4809:.76667] ({x},{.75}) node [pos=0, left] {$-a_6$}; - \addplot[infiniteright,domain=0.4809:.7309] ({x},{.5}) node [pos=1, right] {$a_7$}; + \addplot[firstcurvestyle,infiniteright,domain=0:1] ({x},{2}) node [pos=1, right] {$a_1$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.333:1] ({x},{1.75}) node [pos=0, left] {$-a_2$}; + \addplot[firstcurvestyle,infiniteright,domain=0.333:.833] ({x},{1.5}) node [pos=1, right] {$a_3$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.433:.8333] ({x},{1.25}) node [pos=0, left] {$-a_4$}; + \addplot[firstcurvestyle,infiniteright,domain=0.4333:.76667] ({x},{1}) node [pos=1, right] {$a_5$}; + \addplot[firstcurvestyle,infiniteleft,domain=0.4809:.76667] ({x},{.75}) node [pos=0, left] {$-a_6$}; + \addplot[firstcurvestyle,infiniteright,domain=0.4809:.7309] ({x},{.5}) node [pos=1, right] {$a_7$}; \end{axis} diff --git a/ptx/sec_limit_analytically.ptx b/ptx/sec_limit_analytically.ptx index 9d38baf19..d5ad5b478 100644 --- a/ptx/sec_limit_analytically.ptx +++ b/ptx/sec_limit_analytically.ptx @@ -483,19 +483,19 @@

- Graph illustrating the squeeze theorem. There are three functions, h(x), + An illustration of the squeeze theorem. There are graphs of three functions shown, labelled h(x), g(x), and f(x). On the yaxis, there is a marker at y = 4, labeled L and on the xaxis there is a marker at x = 5, labeled c.

For all values of x \leq c f(x) \leq L and h(x) \geq L. - For all values of x f(x) \leq g(x) \leq h(x), that is, the line - of the function g(x) is between the lines of the functions g(x) + For all values of x f(x) \leq g(x) \leq h(x), that is, the graph + of the function g(x) lies between the graphs of the functions g(x) and f(x).

- The graph shows that where x = c, f(x) and h(x) converge + The image shows that where x = c, f(x) and h(x) converge on y = L = 4. Because f(x) \leq g(x) \leq h(x), we can extrapolate that \lim\limits_{x\to \c} g(x) = L too.

@@ -953,15 +953,15 @@

- Graph of the linear equation (x^1-1)/(x-1), shows the function with - the y interval 0 to 3 and x interval 0 - to 2. The function has a y intercept at y = 1, and - is undefined at x = 1. The graph has a positive slope. + Graph of the function (x^2-1)/(x-1), showing the region with + y from 0 to 3 and x from 0 + to 2. The graph is a straight line, with slope 1 and a y intercept at y = 1, + except for a hole at the point (1,2), since the function is undefined at x = 1.

Graph of the polynomial x squared minus 1 divided by the polynomial x minus 1. - Is undefined at x = 1. + It is the same as the line y=x+1, except that it is undefined at x = 1. \begin{tikzpicture} diff --git a/ptx/sec_limit_continuity.ptx b/ptx/sec_limit_continuity.ptx index 4e7a7b944..909883472 100644 --- a/ptx/sec_limit_continuity.ptx +++ b/ptx/sec_limit_continuity.ptx @@ -105,14 +105,19 @@

- Shows a graph with domain 0 to 3. For 0 \geq x \lt 1 - the graph has a downward curve to it, for 1 > x \leq 2 the graph - is a straight line, parallel with the x axis, and for 2 \geq x \leq 3. - The graph is undefined at x = 1. + The graph of a piecewise-defined function is shown, for x from 0 to 3. + For 0 \leq x \lt 1 + the graph looks like a parabola opening downward. + This part of the graph approaches, but does not reach, the point (1,1). + There is a hollow dot at (1,1), indicating that f(1) is undefined. + For 1 \lt x \leq 2 the graph + is a horizontal line segment, with y=1. + For 2 \leq x \leq 3 the graph again has the appearance of a downward-facing parabola + that begins at (2,1) and ends at (3,1).

- 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. \begin{tikzpicture}[declare function = {func(\x) = (\x >= 0)*(\x <= 1)*(-(\x-1/4)^2+1/16+1.5) + (\x > 1)*(\x <= 2) + (\x > 2)*(\x <= 3)*((2-\x)*(\x-3)+1);}] @@ -202,12 +207,12 @@

- Shows a graph with domain -2 to 3. There are five - straight lines, each parallel with the x 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 (-2, -2), (-1, -2), - line two (-1, -1), (0, -1), line 3 (0, 0), (1, 0), line 4 - (1, 1), (2, 1), and line 5 (2, 2), (3, 2). + Shows the graph of the greatest integer function, for x from -2 to 3. There are five + horizontal line segments in a staircase 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 (-2, -2) to (-1, -2), + the second from (-1, -1) to (0, -1), the third from (0, 0) to (1, 0), + the fourth from (1, 1) to (2, 1), and the fifth from (2, 2) to (3, 2).

@@ -881,17 +886,16 @@

- Shows the graph of a function on the domain 0 to 4. - The graph is has a downward curve - and for x = 2 the function is defined by the point (2, 1). - The point (2, 1) shows a removable discontinuity because the graph - is undefined at x = 2, but the point shows that the function is defined - at x = 2. + A portion of the graph of a function is shown, for x from 0 to 4. + The graph has the shape of a parabola opening downward, + but at x=2 there is a hole in the graph, + and instead the point (2,1) (which is not on the graph) is plotted. + The graph of this function illustrates a removable discontinuity because + \lim_{x\to 2}f(x) exists, but does not equal f(2).

- 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. \begin{tikzpicture} @@ -917,14 +921,21 @@

- Shows the graph of a function on the domain 0 to 4. - When approching x = 2 from the left hand side f(x) is - undefined, but when coming from the right hand side f(x) = 1. For the - interval 0 \lt x \leq 2 the graph is curved downward and for the + The graph of a function is shown for x from 0 to 4. + As x approaches 2 from the left, + the graph of f approaches a point that is not part of the graph, + as indicated by a hollow dot. + As x approaches 2 from the right, + the graph of f 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 x\to 2. +

+ +

+ On the interval 0 \lt x \leq 2 the graph is curved downward and on the interval 2 \leq x \leq 4 the graph is a straight line with a - positive slope. Because f(x) is undefined at x = 2 - when coming from the left, but is defined coming from the right, - there is a jump discontinuity. + positive slope.

@@ -954,15 +965,12 @@

- Shows the graph of a function on the domain 0 to 4. + The graph of a function is shown for x from 0 to 4. There is a - vertical dotted line at x = 2 illustrating an asymptote. - As x approaches 2 from both sides the value of f(x) - approaches infinity. The graph also has an asymptote at y = 0. - On both sides of the dotted vertical line the graph has an upward - curve with an increasing slope at x gets closer to 2. - The asymptote at x = 2 causes there to be an infinite - discontinuity. + vertical dotted line at x = 2 illustrating a vertical asymptote. + As x approaches 2 from either side, + the graph of f extends upward along the asymptote, + indicating that the value of f(x) is increasing without bound.

diff --git a/ptx/sec_limit_infty.ptx b/ptx/sec_limit_infty.ptx index 97c0ca93e..b4182b3f7 100644 --- a/ptx/sec_limit_infty.ptx +++ b/ptx/sec_limit_infty.ptx @@ -37,8 +37,10 @@

Graph of f(x)=1/x^2 for x between -1 and 1. There is a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. - For x values near \infty and -\infty, y - approaches 0. For x values near 0, y approaches \infty. + For x values near the left and right edges of the image, + the y value is close to 0. + For x values near 0, the graph extends to the top of the image (and presumably beyond), + suggesting that y approaches \infty.

@@ -210,9 +212,10 @@ Graph of f(x)=\frac{1}{(x-1)^2} for x between 0 and 1. There is a vertical asymptote at x = 1 and a horizontal asymptote at y = 0. - As x gets near 1 from both sides of the vertical asymptote - y approaches \infty. As x gets near \infty - and -\infty y approaches 0. + As x gets near 1 from either side of the vertical asymptote, + y approaches \infty. + For x values near the left and right edges of the image, + the value of y approaches 0.

@@ -292,8 +295,8 @@ approaches -\infty and from the right, y approaches \infty. As y approaches 0 from the bottom, x approaches -\infty and from the top, x approaches - \infty. The lines of the equation are in quadrants one and - three of the graph. + \infty. The graph conists of two parts; one in quadrant one and the other in + quadrant three.

@@ -476,10 +479,9 @@

- The graph - is a single straight line with a positive slope. At x = 1 - there is a hollow point indicating a discontinuity, the exact - position of the discontinuity is (1, 2). + The graph is a single straight line with a positive slope. + At x = 1 there is a hollow dot indicating a removable discontinuity. + The exact position of the discontinuity is (1, 2).

@@ -680,10 +682,11 @@

- Graph of f(x)=\frac{x^2}{x^2+4} on [-20,20]. There - is a horizontal asymptote at y = 1. As x approaches - -\infty and \infty, f(x) gets near 1, - but never equals 1. The graph drops to the point (0,0) + Graph of f(x)=\frac{x^2}{x^2+4} showing x values from -20 to 20. + There is a horizontal asymptote at y = 1. + As x approaches -\infty and \infty, f(x) gets near 1, + but never equals 1. The graph lies between y=0 and y=1, + and drops to the point (0,0) as x approaches 0 from either direction.

@@ -788,9 +791,8 @@

- As - x approaches -\infty and \infty, f(x) gets near - 0. The graph dips to a minimum value just to the left of the y axis, + As x approaches -\infty and \infty, f(x) gets near 0. + The graph dips to a minimum value just to the left of the y axis, then crosses the x axis at (0,0), rising to a maximum value just to the right of the x axis, before falling again toward the horizontal asymptote y=0.

@@ -821,7 +823,8 @@

The graph of f(x)=\frac{x}{\sqrt{x^2+1}}, which has two horizontal asymptotes, - one at y = -1 and the other at y = 1. + one at y = -1 (representing the limit as x\to -\infty) + and the other at y = 1 (representing the limit as x\to\infty).

@@ -1127,13 +1130,16 @@

- There - is a horizontal asymptote at y = -\frac{1}{3}, helping - illustrate that the limit of f(x) = \frac{1}{3}, - which is the coefficient of the numerator with the highest - power of x in \dfrac{x^2+2x-1}{1-x-3x^2} - divided by the coefficient of the dennominator with the - highest power of x in \dfrac{x^2+2x-1}{1-x-3x^2}. + The graph of f(x)=\frac{x^2+2x-1}{1-x-3x^2} is shown for x\gt 0. + The graph has a horizontal asymptote at y = -\frac{1}{3}, + which it approaches from below. + The graph illustrates that the limit of f(x) as x\to\infty is -\frac{1}{3}. + The coefficient of the term in the numerator of f(x) with the highest + power of x is 1, + while the coefficient of the term in the dennominator with the + highest power of x is -3. + The ratio of these two coefficients gives the limit as x\to\pm \infty + when the highest power of x is the same in both the numerator and the denominator.

@@ -1162,11 +1168,20 @@ +

+ The graph of f(x)=\frac{x^2-1}{3-x} is shown for x\gt 3. + Near x=3 the graph appears to be heading down a vertical asymptote, + suggesting that \lim_{x\to 3^+}f(x)=-\infty. + The graph then rises to a peak, before beginning to descend again. + Beyond x=10, the graph appears almost straight, + and continues downward at a slope close to -1, + showing that there is no horizontal asymptote in this case. +

The graph shows that the limit of f(x) will be determined by dominant terms from the numerator and denominator, which are x^2 and -x. Since \frac{x^2}{-x} = -x for large values - of x, the graph of f(x) behaves approximately the same as that of y=x. + of x, the graph of f(x) behaves approximately the same as that of y=-x.

diff --git a/ptx/sec_limit_intro.ptx b/ptx/sec_limit_intro.ptx index edab7e8ae..b524b5333 100644 --- a/ptx/sec_limit_intro.ptx +++ b/ptx/sec_limit_intro.ptx @@ -38,13 +38,13 @@

- Graph of \sin(x)/x showing the domain and range of - -7 to 7 and 0 to - 1 respectively. The x intercepts are at + Graph of \sin(x)/x, shown for x between + -7 and 7, and y between 0 and + 1. The x intercepts are at x=-2\pi, -\pi, \pi, and 2\pi, and a y intercept is at y = 1. - The function has a downward curve for -\pi \lt x \lt \pi and an - upward curve for -2\pi \gt x \lt -\pi, and - \pi \gt x \lt 2\pi. The graph is undefined for x = 0. + The graph has a downward curve for -\pi \lt x \lt \pi and an + upward curve for -2\pi \lt x \lt -\pi, and + \pi \lt x \lt 2\pi. The graph is undefined for x = 0.

@@ -75,10 +75,10 @@

- Graph of \sin(x)/x zoomed in on values where x is near 1. The - domain of the graph is 0.5 to 1.5. + Graph of \sin(x)/x zoomed in on values where x is near 1. + This view of the graph shows x from 0.5 to 1.5. The graph has only a slight downward curve. It shows that for x = 1, - \sin(x)/x is approx. 0.84 + \sin(x)/x is approximately 0.84

@@ -136,10 +136,11 @@

- Graph of \sin(x)/x zoomed in on values where x is near 0. The - domain of the graph is -1 to 1. The graph - has a downward curve and symmetric, peaking at y = 1. There is a dot at - x = 0 showing the equation is undefined for values of x = 0, + Graph of \sin(x)/x zoomed in on values where x is near 0. + The image shows the portion of the graph where x is from -1 to 1. + The graph has a downward curve and is symmetric about x=0. + The height of the graph approaches y = 1 when x is near 0. + A hollow dot at the point (0,1) shows that the function is undefined when x = 0; that is, f(0) = undefined.

@@ -367,16 +368,19 @@

- Graph of (x^2 - x - 6)/(6x^2 - 19x + 3), - zoomed on values near x = 3. The domain is approximately - 2.5 to 2.5. - There is a slight upward curve to the graph. Shows that the limit of the - equation as x approaches 3 is 0.294. The graph also - shows that the equation is undefined for x = 3. + Graph of f(x)=\frac{x^2 - x - 6}{6x^2 - 19x + 3}, + zoomed on values near x = 3, and showing the portion of the graph + for x from 2.5 to 3.5. +

+

+ There is a slight upward curve to the graph. + The graph suggests that the limit of the function + as x approaches 3 is 0.294. The graph also + shows that the function is undefined for x = 3.

- Graph of the equation, shows that when x = 3 the y = undefined, + Graph of the function for this example, which shows that when x = 3, f(x) is undefined, but near 0.294. @@ -533,9 +537,11 @@

Graph of the piecewise-defined function in . For values of x \lt 0 the graph is straight - with a slope of 1 and for values x \gt 0 the graph is - curved downward with a negative slope. Shows that at x = 0, - y is undefined, but near 1. + with a slope of 1 and for values of x \gt 0 the graph + curves downward. A hollow dot at the point (0,1) shows that at x = 0, + f(x) is undefined. + However, both parts of the graph, for x\lt 0 and for x\gt 0, + get close to the point (0,1) as x gets close to 0.

@@ -697,9 +703,19 @@

Graph of piecewise function in . For values of x \leq 1 the graph - has a upward curve, and where x = 1 y = 2. - For values of x \gt 1 the graph is straight with a - positive slope, where x = 1, y is undefined. + has a upward curve, and the graph ends at the point (1,2), + illustrating the fact that f(1)=2. +

+

+ For values of x \gt 1 the graph is a straight line with a positive slope. + Moving left to right, the line begins at the point (1,1), + at which there is a hollow dot, indicating that to the right of x=1, + the value of f(x) approaches 1. +

+

+ The most important feature of the graph is that it shows how f(x) + approaches two different values as x approaches 1, + depending on whether x\lt 1 or x\gt 1.

@@ -788,12 +804,12 @@ Graph of the function for . The graph hows a horizontal asymptote at y = 0 and a vertical asymptote at x = 1. - Because of the vertical asymptote at x = 1 the equation + Because of the vertical asymptote at x = 1 the function has no limit as x approaches 1.

- Graph of the equation showing that as x approaches 1 f(x) asymptotes. + Graph of the function f(x), showing that as x approaches 1, there is a vertical asymptote. \begin{tikzpicture} @@ -906,16 +922,16 @@

- Graph of sin(1/x) displaying the x and y - intervals -1 to 1. As x gets close to - 0 the cycles shorten, rendering a wide, vertical line. - This line is however not solid as it is just a bunch of lines - really close to each other. + The graph of f(x)=sin(1/x) is shown, for x values between -1 and 1. + Like any sinusoidal graph, the curve oscillates back and forth between y=1 and y=-1. + However, as x gets close to 0, the argument of the sine function increases rapidly, + causing the distance between successive peaks to get smaller and smaller as the graph nears the y axis. + As x gets close to zero, the oscillations get so close together that it is no longer possible to distinguish them, + and the curve appears to become a solid, vertical strip.

- Graph of the equation showing a thick line at x = 0, where the thick - line is just the oscillation of a single thin line on short cycles. + Graph of the function sin(1/x), showing oscillations that become so rapid near the origin that they blur together. \begin{tikzpicture} @@ -979,15 +995,18 @@

- Graph of sin(1/x) displaying the y interval -1 to - 1 and x interval -0.1 to 0.1. As x gets close - to 0 the cycles shorten, rendering a wide, vertical line. - This line is however not solid as it is just a bunch of lines - really close to each other. + Another graph of f(x)=\sin(1/x) is shown, + this time zoomed in to show only the x interval from -0.1 to 0.1. + The features of the graph are the similar to what is visible over the larger interval: + further from the origin, we see the graph oscillating (rapidly) between y=1 and y=-1. + Near the orgin, the oscillations become so rapid that we can no longer tell them apart. + What we conclude from the graph is that on any interval containing x=0, + f(x)=\sin(1/x) takes on every y value between -1 and 1. + (In fact, f(x) attains every value infinitely many times!)

- Graph of the same equation, shown with a smaller x interval, -0.1 to 0.1. + Graph of the same function, sin(1/x), shown with a smaller x interval, -0.1 to 0.1. \begin{tikzpicture} @@ -1175,18 +1194,17 @@

- Graph showing a downward curved equation with the domain and range - to and - to respectively. - There are two dots plotted on the line of the equation at - (1, 10) and (5, 20) with a dotted line intercepting the - points. The dotted line has a positive slope. The line of the - equation intercepts the x axis at (0, 0) + The image shows the graph of a function, along with a line that intersects the graph at two points. + The graph has the shape of a parabola that opens downward, + and is displayed over the region 0\leq x\leq 6, + with a y range from 0 to 25. + There are two points plotted on the graph at coordinates + (1, 10) and (5, 20), and the line through these points is an example of a secant line.

- Graph showing a downward curved equation, with points at (1,10) and (5,20), with a - dotted line intercepting both points and a positive slope. + A downward curved graph, with marked points at (1,10) and (5,20), and a + line intercepting both points. \begin{tikzpicture} @@ -1254,13 +1272,13 @@

- Graph of the same equation from 1.1.26, with the points (1,10) - and (3,21). The secant line has a steeper slope than in - figure 1.1.26. Here the value of h is 2. + Graph of the function from , with the points on the graph (1,10) + and (3,21) marked. A secant line is drawn through these points; it has a steeper slope than in + . Here the value of h is 2.

- Graph of the previous equation, with the points (1,10) + Graph of the same function as the previous figure, with the points (1,10) and (3,21). The secant line has a steeper slope, equal to 5.5. @@ -1287,14 +1305,15 @@

- Graph of the same equation from , but - with the points (1,10) and (2,17). Shows the dotted line with a steeper - slope than in figure 1.1.26. Here the value of h is 1. + Graph of the function from , but + with the points (1,10) and (2,17) on the graph marked. + These points correspond to a value of h=1, + and the secant line through these points has a steeper slope than in .

- Graph of the same equation, with the points (1,10) - and (2,17). The secant line has a steeper slope equal to 7. + Graph of the same function as the previous figure, with the points (1,10) + and (2,17) marked. The secant line has a steeper slope equal to 7. \begin{tikzpicture} @@ -1320,13 +1339,13 @@

- Graph of the same equation from , but - with the points (1,10) and (1.5,13.875). Shows the dotted line with a - steeper slope than in figure 1.1.26. Here the value of h is 0.5. + Graph of the function from , but + with the points (1,10) and (1.5,13.875) on the graph marked, corresponding to the value h=0.5. + The secant line through these points again has a steeper slope than in the previous figures.

- Graph of the same equation, with the points (1,10) and + Graph of the same function, with the points (1,10) and (1.5,13.875). The secant line has a steeper slope equal to 7.75. diff --git a/ptx/sec_limit_onesided.ptx b/ptx/sec_limit_onesided.ptx index dfe88f70b..a34972ca3 100644 --- a/ptx/sec_limit_onesided.ptx +++ b/ptx/sec_limit_onesided.ptx @@ -183,30 +183,31 @@

Graph of the piecewise function f(x) = \begin{cases} x \amp 0\leq x\leq 1 \\ 3-x \amp 1\lt x\lt 2 \end{cases}. - There are two lines, - one with a positive slope and the other with a negative slope. + There are two line segments: for 0\leq x\leq 1 we have a line segment + with positive slope, and for 1\lt x\lt 2 we have a line segment with negative slope.

- The line with a positive slope starts at the point (0, 0) - and ends at (1, 1). The line with a negative slope starts + The line segment with a positive slope starts at the point (0, 0) + and ends at (1, 1). The line segment with a negative slope starts at (1, 2) and ends at (2, 1).

- The start and end points of the line with a positive slope - are solid dots, indicating that the function is defined - at those points. The start and end points of the line - with a negative slope are hollow dots. This tells us that - the function is undefined for x = 1 when going right - to left and x = 2 is undedined as well. Because the - function is defined at x = 1 on the line with a positive - slope, but is undefined at x = 1 on the other line, we can - tell the function f has a one-sided limit as x - approaches 1. + The start and end points of the line segment with a positive slope + are solid dots, indicating that those points are part of the graph. + The start and end points of the line segment + with a negative slope are hollow dots. + This tells that although the second line segment gets arbitrarily close + to the points (1,2) and (2,1), these points are not part of the graph. +

+ +

+ Since f(x) is close to 1 when x is close to 1, but x\lt 1, + while f(x) is close to 2 when x is close to 1, but x\gt 1, + we can conclude that the left and right hand limits are different.

- Graph of a piecewise function that has only a left sided limit as x - approaches 1. + Graph of a piecewise function that has different left and right hand limits when x=1. \begin{tikzpicture} @@ -532,23 +533,22 @@

Graph of f from . - There are - two lines shown, - the first starts at the point 0, 2 and ends at - (1, 1). The second start at (1, 1) and ends - at (2, 0). Both lines have a negative slope and are - undefined at their start and end points, which are marked with - hollow dots. The first line is straight, but the second has a - small upward curve to it. + The graph consists of two parts. + The first part is a line that starts at the point (0, 2) and ends at + (1, 1). The second part is a curve that starts at (1, 1) and ends + at (2, 0). + The points (0,2), (1,1), and (2,0) are all marked with hollow dots, + indicating that although the graph gets close to these points, they are not part of the graph.

- The graph may be undefined for x = 1, but since - the equation has a left and right limit, there for - f(x) has a limit as x approaches 1. + The function is undefined for x = 1, + but the graph shows that f(x) approaches the same value (namely, 1) + from both the left and the right, allowing us to conclude that \lim_{x\to 1}f(x) exists, + and is equal to 1.

- Graph of the piecewise function from the previous example. + Graph of a piecewise-defined function. It is undefined when x=1, but has a limit at this point. \begin{tikzpicture}[declare function = {func(\x) = (\x < 1) * (2 - x) + (\x > = 1) * ((x - 2)^2);}]