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.