Applications-of-GeoGebra/C3/Limits-and-Continuity-of-Functions/English

Visual Cue	Narration
Slide Number 1 Title Slide	Welcome to this tutorial on Limits and Continuity of Functions.
Slide Number 2 Learning Objectives	In this tutorial, we will learn how to use GeoGebra to: Understand limits of functions Look at continuity of functions
Slide Number 3 System Requirement	Here I am using: Ubuntu Linux OS version 16.04 GeoGebra 5.0.481.0-d
Slide Number 4 Pre-requisites www.spoken-tutorial.org	To follow this tutorial, you should be familiar with: GeoGebra interface Limits Elementary calculus For relevant tutorials, please visit our website.
Slide Number 5 Limits [[Image:]][[Image:]] [[Image:]]	Let us understand the concept of limits by looking at three graphs A, B and C. Imagine yourself sliding along the curve or line towards a given value of x. The height at which you will be, is the corresponding y value of the function. Any value of x can be approached from two sides. The left side gives the left hand limit. The right side gives the right hand limit.
Slide Number 6 Left hand and right hand limits [[Image:]] lim_(x→b) f(x) = ? lim_(x→ b-) f(x) = L1; lim_(x→b+) f(x) = L1 = f(b)	In graph A, let us find the limit of f of x as x approaches or tends to b. f of x is a continuous line. The left hand limit of f of x as x tends to b is L1. And the right hand limit of f of x as x tends to b is also L1. Thus, the limit of f of x as x approaches b is L1. It is the same as evaluating f of x at x equals b, that is, f of b.
Slide Number 7 Left hand and right hand limits [[Image:]] lim_(x→b1) g(x) =? lim_(x→b1-) g(x) = lim_(x→b1+) g(x) = L2' But g(b1) does not exist (DNE) lim_(x→b) g(x) = g(b) = L2; lim_(x→a) g(x) = g(a) = L1	What is the limit of g of x as x tends to b1? In graph A, note that g of x has an open circle at b1 comma L2 prime. This means that g of x does not exist at this point. Let us find the limit of g of x as x approaches b1. The left hand and right hand limits are L2' as x approaches b1. But g of x itself does not exist at x equals b1. However, g of x can be evaluated at x equals b and x equals a. And these values are the same as the limits of g of x as x approaches b and a.
Slide Number 8 Limits of discontinuous functions [[Image:]] lim_(x→c) h(x) = ? lim_(x→c-) h(x) = L4; lim_(x→c+) h(x) = L3 Thus, lim_(x→c) h(x) does not exist (DNE)	In graph B, h of x is a piecewise or discontinuous function. We want to find the limit of h of x as x approaches c. So let us look at the left and right hand limits. For the left hand limit, look at the lower limb where the limit is L4. For the right hand limit, look at the upper limb where limit of h of x is L3. But as x approaches c, the two limbs of h of x approach different values of y. These are L3 and L4. The left and right hand limits exist. But the limit of h of x as x approaches c, does not exist (DNE).
Slide Number 9 Criteria for continuous functions Note that for a function f(x) to be called continuous, f(a) should be defined lim_(x→ a) f(x) exists lim_(x→ a) f(x) = f(a)	Note that for a function f of x to be called continuous, f of a should be defined limit of f of x as x tends to a exists limit of f of x as x tends to a is f of a
Slide Number 10 Limits at infinity [[Image:]] lim_(x→∞) i(x) = ? lim_(x→-∞) i(x) = ? lim_(x→∞) i(x) = 2; lim_(x→-∞) i(x) = 1	In graph C, i of x has two parts. The first part is the upper right one. Both arms extend towards infinity (∞). The second part is the lower left one. Both arms extend towards negative infinity (-∞). What are the limits of i of x as x tends to infinity and minus infinity? The limit of i of x as x approaches infinity is 2. And the limit of i of x as x approaches negative infinity is 1.
Slide Number 11 Limit of a rational polynomial function Let us find lim_(x→2) (3x2 – x -10)/(x2 – 4)	Let us find the limit of this rational polynomial function as x tends to 2.
Show the GeoGebra window.	I have already opened the GeoGebra interface. Let us graph functions and look at their limits.
To type the caret symbol, hold the Shift key down and press 6. Type (3 x2-x-10)/(x2-4) in the input bar >> Enter	To type the caret symbol, hold the Shift key down and press 6. Note that spaces denote multiplication. In the input bar, first type the numerator. In parentheses, 3 space x caret 2 minus x minus 10 followed by division slash Now, type the denominator. In parentheses, x caret 2 minus 4 Press Enter.
Point to the equation in Algebra view and its graph in Graphics view.	The equation appears in Algebra view and its graph in Graphics view.
Click on Move Graphics View tool. Click in and drag Graphics view to see the graph.	Click on Move Graphics View. Click in and drag Graphics view to see the graph.
Point to the graph in Graphics view.	As x approaches 2, the function approaches some value close to 3.
Click on View tool and select Spreadsheet.	Click on View and select Spreadsheet.
Point to the spreadsheet on the right side of the Graphics view.	This opens a spreadsheet on the right side of the Graphics view.
Click on Options tool and click on Rounding and choose 5 decimal places.	Click on Options and click on Rounding and choose 5 decimal places.
Remember to press Enter to go to the next cell. Type 1.91, 1.93, 1.96, 1.98 and 2 in column A from cells 1 to 5.	Let us find the left hand limit of this function as x tends to 2. We will choose values of x less than but close to 2. Remember to press Enter to go to the next cell. In column A in cells 1 to 5, type 1.91, 1.93, 1.96, 1.98 and 2.
Type 2.01, 2.03, 2.05, 2.07 and 2.09 in column A from cells 6 to 10.	Let us find the right hand limit of this function as x tends to 2. We will choose values of x greater than but close to 2. In column A from cells 6 to 10, type 2.01, 2.03, 2.05, 2.07 and 2.09.
In cell B1 (that is, column B, cell 1), type (3(A1)^2-A1-10)/((A1)^2-4) >> Enter.	In cell B1 (that is, column B, cell 1), type the following ratio of values. First, the numerator in parentheses 3 A1 in parentheses caret 2 minus A1 minus 10 followed by division slash Now the denominator in parentheses A1 in parentheses caret 2 minus 4 and press Enter.
Click on cell B1 to highlight it. Place the cursor at the bottom right corner of the cell. Drag the cursor to highlight cells until B10. Point to y values in column B and to the x values in column A.	Click on cell B1 to highlight it. Place the cursor at the bottom right corner of the cell. Drag the cursor to highlight cells until B10. This fills in y values corresponding to the x values in column A.
Drag and increase column width.	Drag and increase column width.
Point to the question mark in cell B5 corresponding to x=2. Point to the spreadsheet.	Note that a question mark appears in cell B5 corresponding to x equals 2. This is because the function is undefined at this value. The reason for this is that the denominator of the function becomes 0. Observe that as x tends to 2, y tends to 2.75. Hence, as x tends to 2, the limit of the function tends to 2.75.
Slide Number 12 Limit of a rational polynomial function lim_(x→2) (3x2 – x -10)/(x2 – 4) = 2.75	Thus, the limit of this rational polynomial function as x tends to 2 is 2.75.
Slide Number 13 Limit of a discontinuous function Let us find lim_(x→0) f(x) = 2x+3, x ≤ 0 ................................ =3(x+1), x > 0 and lim_(x→1) f(x) = 2x+3, x ≤ 0 ...........................= 3(x+1), x > 0	Let us find limits of a piecewise or discontinuous function f of x. f of x is described by 2x plus 3 when x is 0 or less than 0. But f of x is described by 3 times x plus 1 when x is greater than 0. We want to find the limits when x tends to 0 and 1.
Open a new GeoGebra window.	Let us open a new GeoGebra window.
Type a=Function[2x+3,-5,0] in the input bar >> Enter	In the input bar, type the following line. a equals Function with capital F and in square brackets 2x plus 3 comma minus 5 comma 0 This chooses the domain of x from minus 5 (for practical purposes) to 0. Press Enter.
Point to the equation a(x)=2x+3 (-5 ≤ x ≤ 0) in Algebra view. Drag the boundary to see it properly. Point to its graph in Graphics view.	The equation a of x equals 2x plus 3 where x varies from minus 5 to 0 appears in Algebra view. Drag the boundary to see it properly. Its graph is seen in Graphics view.
Under Move Graphics View, click on Zoom Out and click in Graphics view.	Under Move Graphics View, click on Zoom Out and click in Graphics view.
Click on Move Graphics View and drag the background to see the graph properly.	Click on Move Graphics View and drag the background to see the graph properly.
Click on Move Graphics View tool, place cursor on x-axis. When an arrow appears along the axis, drag the x-axis to zoom in or out.	Click on Move Graphics View and place the cursor on the x-'axis. When an arrow appears along the axis, drag the x-axis to zoom in or out.
Similarly, click on Move Graphics View tool and place cursor on y-axis. When an arrow appears along the axis, drag the y-axis to zoom in or out.	Similarly, click on Move Graphics View and place the cursor on the y-axis. When an arrow appears along the axis, drag the y-axis to zoom in or out.
Click in and drag the background to see the graph properly.	Click in and drag the background to see the graph properly.
Type b=Function[3(x+1),0.01,5] in the input bar >> Enter	In the input bar, type the following command and press Enter. Remember the space denotes multiplication. b equals Function with capital F In square brackets, type 3 space x plus 1 in parentheses comma 0.01 comma 5 This chooses the domain of x from 5 (for practical purposes) to 0.01. For this piece of the function, x is greater than 0 but not equal to 0.
Point to the equation b(x)=3(x+1) (0.01 ≤ x ≤ 5) in Algebra view. Point to its graph in Graphics view.	The equation b of x equals 3 times x plus 1 where x varies from 0.01 to 5 appears in Algebra view. Its graph is seen in Graphics view.
Double click on the equation a(x)=2x+3 in Algebra view.	In Algebra view, double click on the equation b of x equals 3 times x plus 1.
Click on Object Properties.	Click on Object Properties.
Click on Color tab and select blue.	Click on the Color tab and select blue.
Close the Preferences dialog box.	Close the Preferences dialog box.
	Click in and drag the background to see both functions in Graphics view.
	Under Move Graphics View, click on Zoom In. Now click on Move Graphics View and drag the background until you can see both graphs.
Point to the break between the blue and red functions for f(x)=3(x+1).	Note that there is a break between the blue and red functions. This is because x is not 0 when f of x is 3 times x plus 1.
Point to the blue function. Point to intersection of f(x) and y-axis at (0,3).	The blue function has to be considered for x less than and equal to 0. When x tends to 0, f of x is 3 as the function intersects the y-axis at 0 comma 3.
Point to the red function.	The red function has to be considered for x greater than 0. When x equals 1, the value of f of x is 6.
Slide Number 14 Limit of a discontinuous function lim_(x→0) f(x) = 2x+3, x ≤ 0 }=3 ..........................= 3(x+1), x > 0 and lim_(x→1) f(x) = 2x+3, x ≤ 0 }=6 .........................= 3(x+1), x > 0	Thus, for this discontinuous function, f of x is 3 when x is 0. When x is 1, f of x is 6.
	Let us summarize.
Slide Number 15 Summary	In this tutorial, we have learnt how to use GeoGebra to: Understand limits of functions Look at continuity of functions
Slide Number 16 Assignment Find the limit of (x3-2x2)/(x2-5x+6) as x tends to 2. Evaluate lim_(x→0) sin 4x/sin 2x	As an Assignment: Find the limit of this rational polynomial function as x tends to 2. Find the limit of this trigonometric function as x tends to 0.
Slide Number 17 About Spoken Tutorial project	The video at the following link summarizes the Spoken Tutorial project. Please download and watch it.
Slide Number 18 Spoken Tutorial workshops	The Spoken Tutorial Project team: * conducts workshops using spoken tutorials and * gives certificates on passing online tests. For more details, please write to us.
Slide Number 19 Forum for specific questions: Do you have questions in THIS Spoken Tutorial? Please visit this site Choose the minute and second where you have the question Explain your question briefly Someone from our team will answer them	Please post your timed queries on this forum.
Slide Number 20 Acknowledgement	Spoken Tutorial Project is funded by NMEICT, MHRD, Government of India. More information on this mission is available at this link.
	This is Vidhya Iyer from IIT Bombay, signing off. Thank you for joining.

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Madhurig, Vidhya