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.