Basic Calculus - Q3 Module 1 PDF

11 SENIOR HIGH SCHOOL BASIC CALCULUS Quarter 3 – Module 1 The Limit of a Function and Limit Laws  NegOr_Q3_Basic Calculus11_Module1_v2 NegOr_Q3_Basic Calculus11_Module1_v2 Basic Calculus – Grade 11...

11 SENIOR HIGH SCHOOL BASIC CALCULUS Quarter 3 – Module 1 The Limit of a Function and Limit Laws  NegOr_Q3_Basic Calculus11_Module1_v2 NegOr_Q3_Basic Calculus11_Module1_v2 Basic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 1: The Limit of a Function and Limit Laws Second Edition, 2021 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio Development Team of the Module Writer: Mercyditha D. Enolpe Editors: Ronald G. Tolentino & Gil S. Dael Reviewer: Littie Beth S. Bernadez  Layout Artist: Radhiya A. Ababon Management Team: Senen Priscillo P. Paulin CESO V Elisa L. Bagiuo EdD Joelyza M. Arcilla EdD, CESE Rosela R. Abiera Marcelo K. Palispis JD, EdD Maricel S. Rasid Nilita L. Ragay EdD Elmar L. Cabrera Printed in the Philippines by ________________________ Department of Education –Region VII Schools Division of Negros Oriental Office Address: Kagawasan, Ave., Daro, Dumaguete City, Negros Oriental Tel #: (035) 225 2376 / 541 1117 E-mail Address: [email protected] NegOr_Q3_Basic Calculus11_Module1_v2 Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson. Each SLM is composed of different parts. Each part shall guide you step-by step as you discover and understand the lesson prepared for you. Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these. In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning. Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task. If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Thank you.  i NegOr_Q3_Basic Calculus11_Module1_v2 I This module was designed to provide you with fun and meaningful opportunities for guided and independent learning at your own pace and time. You will be enabled to process the contents of the learning resource while being an active learner. The module is intended for you to illustrate the limit of a function using table of values and the graph of a function and the limit laws; distinguish between lim → ( ) ( ); and apply the limits laws in evaluating the limit of algebraic functions (polynomial, rational, and radical) I PRE-ASSESSMENT Multiple Choice. Read and analyze each item carefully and write the letter of the correct answer on your activity sheets/notebook. 1. Given the table of values below. Which among the statements DOES NOT correctly describe the values in the table? x f(x) x f(x) 0.5 -3.5 1.5 -2.5 0.88 -3.12 1.17 -2.83 0.996 -3.004 1.003 -2.997  0.999 -3.0001 1.0001 -2.9999 9 A. The table shows that as the value of x increases the value of f(x) also increases. B. The table shows that as the value of x decreases the value of f(x) also decreases. C. The table shows that as x approaches 1 from left or right, f(x) approaches -3. D. The table shows that as the value of x decreases the value of f(x) increases. 1 NegOr_Q3_Basic Calculus11_Module1_v2 For items 2-4, refer to the given. Let ( ) = 3−3 2+ −3 −3. 2. Given the table of values below, what is the value of f(x) if x = 2.99999? x f(x) 2.7 8.29 2.85 9.12250000000001 2.995 9.97002499999939 2.99999 A. 6.0111 B. 9.99993876 C. 10.0001 D. 10.51 3. Given the table of values below, what is the value of f(x) if x = 3.00001? x f(x) 3.5 13.25 3.1 10.61 3.001 10.006000999997 3.00001 A. 10.00006 B. 10.51 C. 13.25 D. 13.99 4. What do the tables in items 2 and 3 show based on their limit values? A. The tables show that as x approaches 3 from the left or right, f(x) approaches to the same value of 10. B. The tables show that as x approaches -3 from the left and 3 from the right, f(x) approaches to the same value. C. The tables show that as the value of x increases, f(x) decreases. D. The tables show that as the value of x decreases, f(x) increases. 5. Does the limit exist for this function as x approaches to 3?  A. No, only left hand limit exist. C. Yes. Both hand limit exist.. B. No, only right hand limit exist. D. No limit exist. 6. Which of the following expressions follow the limit of a constant? A. lim →2 = 2 C. lim →2 2 = 4 B. lim → −3.14 = , ∈ ℝ D. lim → 7 = 7, ∈ ℝ 7. Which of the following illustrates the Multiplication Theorem? A. lim → ( ( )) = lim → ( ( )) =. B. lim → ( ) = C. lim → ( ( ) ∙ ( )) = lim → ( ) ∙ lim → ( ) = ∙. D. lim → ∙ ( ) = ∙ lim → ( ) = ∙. 8. What theorem states that the limit of a sum/difference of functions is the sum/difference of the limits of the individual functions? A. The Summing Up Theorem C. The Addition/Subtraction Theorem B. Limitless Theorem D. The Add On Theorem 2 NegOr_Q3_Basic Calculus11_Module1_v2 9. How is the Division Theorem written in symbols if it states that the limit of a quotient of functions is equal to the quotient of the limits of the individual functions, provided the denominator limit is not equal to 0? lim ( )= → ( ) A. lim → ( ) , provided M = lim → ( )= 0. lim ( )= → ( ) B. lim → ( ) , provided M > lim → ( )= 0. lim ( )= → ( ) C. lim → ( ) , provided M < lim → ( )= 0. lim ( )= → ( ) D. lim → ( ) = provided M ≠ 0. lim → ( )= , √lim → ( ) , in words it states that: 10. The Radical/ Root = √ Theorem lim → √ ( ) A. If n is a negative integer, the limit of the nth root of a function is just the nth root of the limit of the function, provided the nth root of the limit is a real number. Thus, it is important to keep in mind that if n is even, the limit of the function must be positive. B. If n is a positive integer, the limit of the nth root of a function is just the nth root of the limit of the function, provided the nth root of the limit is a real number. Thus, it is important to keep in mind that if n is even, the limit of the function must be positive. C. If n is a positive integer, the limit of the nth root of a function is just the limit of the function, provided the nth root of the limit is a negative number. D. If n is a even integer, the limit of the nth root of a function is just the nth root of the limit of the function, provided the nth root of the limit is a real number. Thus, it is important to keep in mind that if n is odd, the limit of the function must be negative.  3 NegOr_Q3_Basic Calculus11_Module1_v2 Lesson 1 Values and Graph and Distinction between → ( ) and f(c) lim Illustration of Limit of a Function using the Table of ’s In PRIOR-KNOWLEDGE Complete the table of values below given the functions f(x) = 1 + 3x x -3 -2 -1 0 1 2 3 f(x) and g(x) = x2– 1 x -3 -2 -1 0 1 2 3 f(x) Now using the table above, sketch the graph of the two functions.  4 NegOr_Q3_Basic Calculus11_Module1_v2 From the graph of the two functions as shown at the left, how will you reflect on the g(x) = x2– 1 following? a. What happened to f(x) and g(x) if x increases in value approaching 0? What value is the function approaching then? ______________________ b. What happened to the function as the value of x decreases approaching 0? What value is it approaching then? ______________________ c. Are the two values similar? So, what do f(x) = 2x + 5 you call this value of f(x)? Figure 1. Graph of f(x) and g(x) ______________________ ’s New Activity 1. Try This! Evaluate the following functions. You can refer to figure 1 above for the graph of the functions f(x) and g(x). 1. f(x) = 1 + 3x, = 2 2. g(x) = x2 − 1, = 2  5 NegOr_Q3_Basic Calculus11_Module1_v2 is It 1.1. ILLUSTRATING THE LIMIT OF A FUNCTION USING TABLE OF VALUES AND GRAPH Consider a function f of a single variable x. Consider a constant c which the variable x will approach (c may or may not be in the domain of f). The limit, to be denoted by L, is the unique real value that f(x) will approach as x approaches c. In symbols, we write this process as → ( ) =. This is read as, “The limit of f(x) as x approaches to c is L.” To illustrate, let us consider lim →2(1 + 3 ). Here, ( ) = (1 + 3 ) and the constant c, which x will approach is 2. To evaluate the given limit, we will make use of the table to help us keep track of the effect that the approach of x toward 2 will have on f(x). x f(x) 1 4 1.4 5.2 1.7 6.1 1.9 6.7 1.95 6.85 1.997 6.991 1.9999 6.9997 1.9999999 6.9999997 Now, consider approaching 2 from its right or through values greater than but close to 2. x f(x) 3 10  2.5 8.5 2.2 7.6 2.1 7.3 2.03 7.09 2.009 7.027 2.0005 7.0015 2.0000001 7.0000003 Observe that as the values of x get closer and closer to 2, the values of f(x) get closer and closer to 7. This behavior can be shown no matter what set of values, or what direction, is taken in approaching 2. In symbols, it is written as lim →2(1 + 3 )= 7. 6 NegOr_Q3_Basic Calculus11_Module1_v2 Consider again f(x) = 1+3x. Its graph is the straight line with slope 3 and intercepts (0,1) and (−13, 0). Look at the graph in the vicinity of x = 2. You can easily see the points (from the table of values above (1,4), (1.4, 5.2), (1.7,6.1), and so on, approaching the level where f(x) = 7. The same can be seen from the right. Hence, the graph clearly confirms that lim →2(1 + 3 )= 7. y = 1 + 3 (2, 7) x Example 1. Investigate limx→−1(x2 + 1) by constructing tables of values. Here, c = -1 and f(x) = 2 + 1. We start again by approaching -1 from the left. x f(x) -1.5 3.25 -1.2 2.44 -1.01 2.0201  -1.0001 2.00020001 Now, approach -1 from the right x f(x) -0.5 1.25 -0.8 1.64 -0.99 1.9801 -0.9999 1.99980001 The tables show that as x approaches -1, f(x) approaches 2. Therefore, limx→−1(x2 + 1) = 2. 7 NegOr_Q3_Basic Calculus11_Module1_v2 2 The graph of f(x)= x + 1, is given below: It can be seen from the graph that as values of x approach to -1, the values of f(x) approach 2. Example 2. Investigate limx→0| | through a table of values. Approaching 0 from the left and from the right, we get the following tables: x | | x | | -0.3 0.3 0.3 0.3 -0.01 0.01 0.01 0.01 -0.00009 0.00009 0.00009 0.00009 -0.00000001 0.00000001 0.00000001 0.00000001 Hence, limx→0| | = 0. In this example, f(x) = | |. The graph is presented below:  It can be seen from the graph that as values of x approach to 0, the values of f(x) approach 0 also. 8 NegOr_Q3_Basic Calculus11_Module1_v2 Example 3. Investigate lim →1( 2−5 +4 −1) by constructing tables of values. Here, c = 1 and 2 f(x) = −5 +4 −1.Take note that 1 is not in the domain of f, but this is not a problem. In evaluating a limit, remember that we only need to go very close to 1, we will not go to 1 itself. We now approach 1 from left. x f(x) 1.5 -2.5 1.17 -2.83 1.003 -2.997 1.0001 -2.9999 Then, we approach 1 from the right. x f(x) 0.5 -3.5 0.88 -3.12 0.996 -3.004 0.9999 -3.0001 The tables show that as x approaches to 1, f(x) approaches to -3. In symbols, lim →1( 2−5 +4 −1) = -3. Since f(x) = 2−5 +4 −1, the graph shows this way  Take note that f(x) = 2−5 +4 ( −4)( −1) −1, then f(x) = −1. Thus, f(x) = x - 4, provided x ≠ 1.Hence, the graph of f(x) is also the graph of y = x-1, excluding the point where x = 1. 9 NegOr_Q3_Basic Calculus11_Module1_v2 1.2. DISTINGUISHING BETWEEN → ( ) AND ( ) Is lim → ( ) always equal to ( )? To answer this question, consider the table of values of the function ( ) below. The table of values as x approaches to the left or to the right of 2 is presented below. x f(x) 1 4 1.4 5.2 1.7 6.1 1.9 6.7 1.95 6.85 1.997 6.991 1.9999 6.9997 1.9999999 6.9999997 x f(x) 3 10 2.5 8.5 2.2 7.6 2.1 7.3  2.03 7.09 2.009 7.027 2.0005 7.0015 2.00000001 7.00000003 We can conclude that lim →2(1 + 3 ) = 7. While, (2) = 7. So, in this example, lim →2 ( ) (2) are equal. Notice that the same holds for the following examples as discussed. → f(c) ( ) limx→−1x2 + 1 = 2 f(-1)=2 lim →0| |=0 f(0)=0 This, however, is not always the case. Let us consider the function. f(x) = {| | ≠ 0 2 ≠ 0 Does this in any way affect the existence of the limit? Not at all. This example shows that → ( ) and f(c) may be distinct. 10 NegOr_Q3_Basic Calculus11_Module1_v2 Furthermore, consider the third example where f(x) = { + 1 < 4 ( − 4)2 + 3 ≥4 We have, lim → ( f(c) ) lim → ( f(4)=2 ) Once again, we see, that → ( ) and f(c) are not the same. ’s More Activity Complete the following tables to investigate lim →1( 2 − 2 + 4). x f(x) 0.5 0.7 0.95 0.995 0.9995 0.99995 x f(x) 1.6 1.35 1.05 1.005 1.0005 1.00005 Answer the following questions. 1. What observations do you get from the result in the activity above? 2. Given the table of values. What does this show base on their limit values? 3. Describe the values of x in the two tables. 11 NegOr_Q3_Basic Calculus11_Module1_v2 Lesson 2 and their Applications in Evaluating the Limit of Functions Illustration of Limit Laws ’s In PRIOR-KNOWLEDGE  Lesson 1 showed us how limits can be determined through either a table of values or the graph of a function. One might ask: Must one always construct a table or graph the function to determine the limit? Filling in a table of values sometimes requires very tedious calculations. Likewise, a graph may be difficult to sketch. However, these should not be reasons for a student to fail to determine a limit. In this lesson, we will learn how to compute the limit of a function using Limit Theorems. ’s New Activity 1: Construct the table of values to evaluate the following: 1. lim →1 = 2. lim →2 2 = 3. lim →4√ = 4. lim →−25 = 12 NegOr_Q3_Basic Calculus11_Module1_v2 is It 2.1 ILLUSTRATION OF LIMIT LAWS The limit laws enable us to directly evaluate limits, without need for a table or a graph. Let c is a constant, and f and g are functions which may or may not have c in their domains. LIMIT OF A CONSTANT i. The limit of a constant is itself. If k is any constant, then, lim → =. Example: 1.a. lim → 2 = 2 1.b. lim → −3.14 = −3.14 1.c. lim → 789 = 789 ii. The limit of x as x approaches c is equal to c. This may be thought of as the  substitution law because x is simply substituted by c. lim → = Example: 2.a. lim →9 = 9 2.b. lim →0.005 = 0.005 2.c. lim →−10 = −10 For the remaining theorems, we will assume that the limits of f and g both exist as x approaches c and that they are L and M, respectively. In other words, lim → ( ) = , and lim → ( ) = iii. The Constant Multiple Theorem: This says that the limit of a multiple of a function is simply that multiple of the limit of the function. lim → ∙ ( ) = ∙ lim → ( ) = ∙. For example, let lim → ( ) = 4. Then a. lim → 8 ∙ ( ) = 8 ∙ lim → ( ) = 8 ∙ 4 = 32. b. lim → −11 ∙ ( ) = −11 ∙ lim → ( ) = −11 ∙ 4 = −44. c. lim → 32∙ ( ) =32∙ lim → ( ) =32∙ 4 = 6. 13 NegOr_Q3_Basic Calculus11_Module1_v2 iv. The Addition Theorem: This says that the limit of a sum of functions is the sum of the limits of the individual functions. Subtraction is also included in this law, that is, the limit of a difference of functions is the difference of their limits. In symbols, lim → ( ( ) + ( )) = lim → ( ) + lim → ( ) = +. lim → ( ( ) − ( )) = lim → ( ) − lim → ( ) = −. For example, if lim → ( ) = 4 and lim → ( ) = −5, then a. lim → ( ( ) + ( )) = lim → ( ) + lim → ( ) = 4 + (−5) = −1. b. lim → ( ( ) − ( )) = lim → ( ) − lim → ( ) = 4 − (−5) = 9. v. The Multiplication Theorem: This is similar to the Addition Theorem, with multiplication replacing addition as the operation involved. Thus, the limit of a product of functions is equal to the product of their limits. In symbols, lim → ( ( ) ∙ ( )) = lim → ( ) ∙ lim → ( ) = ∙. For example, let lim → ( )=4 and lim → ( ) = −5. Then lim → ( ( ) ∙ ( )) = lim → ( ) ∙ lim → ( ) = 4 ∙ (−5) = −20. v.a. Remark 1: The Addition and Multiplication Theorems may be applied to sums,  differences and products of more than functions. Remark 2: The Constant Multiple Theorem is a special case of the Multiplication Theorem. Indeed, in the Multiplication Theorem, if the first function f(x) is replaced by a constant k, the result is the Constant Multiple Theorem. 14 NegOr_Q3_Basic Calculus11_Module1_v2 vi. The Division Theorem: This says that the limit of a quotient of functions is equal to the quotient of the limits of the individual functions, provided the denominator limit is not equal to 0. In symbols, lim ( )= → ( ) lim → ( ) lim → ( )= , provided M ≠ 0. For example, a. If lim → ( ) = 4 and lim → ( ) = −5 lim ( )= → ( ) lim → ( ) lim → ( = = −45. ) b. If lim → ( ) = 0 and lim → ( ) = −5 lim → ( ) = −05= 0. lim → ( )= c. If lim → ( ) = 4 and lim → ( ) = 0, it is not possible to evaluate lim → ( ) ( ), or we may say that the limit DNE (does not exist). vii. The Power Limit. This theorem states that the limit of an integer power p of a function is just that power of the limit of a function. In symbols, lim → ( ( )) = For example, lim → ( ( )) = a. If lim → ( ) = 4, then  lim → ( ( ))3 = 43 = 64. b. If lim → ( ) = 4, then 2 lim → ( ( ))−2 = 4−2 = 14 = 116. viii. The Radical/ Root Theorem. This theorem states that if n is a positive integer, the limit of the nth root of a function is just the nth root of the limit of the function, provided the nth root of the limit is a real number. Thus, it is important to keep in mind that if n is even, the limit of the function must be positive. In symbols, = √lim → ( ) lim → √ ( ) For example, = √. = 2. = √lim → ( ) a. If lim → ( )=4, then lim → √ ( ) because b. If lim → ( )=-4, then it is not possible to evaluate lim → √ ( ) and this is not a real number. lim → √ ( ) 15 NegOr_Q3_Basic Calculus11_Module1_v2 2.2 APPLICATION OF LIMIT LAWS IN EVALUATING THE LIMIT OF ALGEBRAIC FUNCTIONS Recall the limit theorems. These theorems will be used in evaluating algebraic functions and illustrated in the following examples. Limits of Polynomial Functions We start with evaluating of polynomial functions. Example 1. Determine lim →1(2 + 1) Solution: From the theorems above, lim →1(2 + 1) = lim →12 + lim →11 (Addition) = (2lim →1 ) + 1 (Constant Multiple) = 2(1) + 1 (lim → = ) =2+1 = 3. Example 2. Determine lim →−1(2 3 − 4 2 + 1) Solution: From the theorems above, lim →−1(2 − 4 2 + 1) = lim →−12 3 − lim →−14 2+ lim →−11 (Addition) 3 = 2lim →−1 3 − 4 lim →−1 2 (Constant Multiple) =2(−1)3 – 4(−1)2+ 1 (Power) = -2-4+1 = -5. Example 3. Evaluate lim →0(3 4 − 2 − 1). Solution. From the theorems above, lim →0(3 4 − 2 − 1) =lim →03 4- lim →0(2 -lim →01 (Addition) =3 lim →0 4- 2lim →0( -1 (Constant Multiple) = 3(0)4-2(0) – 1 (Power)  = 0 – 0 -1 = -1 Limits of Rational Functions We will now apply the limit theorems in evaluating rational functions. In evaluating the limits of such functions, recall from Theorem 1 the Division Rule, and all the rules stated in Theorem 1 which have been useful in evaluating limits of polynomial functions, such as the Additional and Product Rules. Example 4. Evaluate lim →11. Solution: First, note that lim →1 = 1. Since the limit of the denominator is nonzero, we can apply the Division Rule. Thus, lim →11 =lim →11 lim →1 (Division) = 11 = 1. 16 NegOr_Q3_Basic Calculus11_Module1_v2 Example 5. Evaluate lim →2 +1 Solution: Start checking the limit of the polynomial function in the denominator. lim →2( − 1) = lim →2 − lim →21 =2-1 =1 Since the limit of the denominator is not zero, it follows that lim − 1= →2 lim →2 = 21 (Division) lim →2( − 1) =2 Example 6. Evaluate lim →1( −3)( 2−2) +1. First, note that 2 lim →1( 2 + 1) = lim →1 2 +lim →11 =1 + 1 = 2 ≠ 0. Thus, using the theorem, lim +1= 2 2 →1( −3)( −2) lim →1( −3)( 2−2) lim →1( +1)(Division) 2 = lim →1( −3)∙lim →1( 2−2) 2 (Multiplication) (lim = 2 →1 −lim →13)(lim →1 −lim →12) 2 (Addition) (1−3)(12 = −2) 2 = 1. Limits of Radical Functions We will now evaluate limits of radical functions using limit theorems. Example 7. Evaluate lim →1√. Solution. Note that lim →1 = 1 > 0. Therefore, by the = radical/root rule, lim →1√  √lim →1 =√1=1 Example 8. Evaluate lim →0√ + 4. Solution. Note that lim →0( + 4) = 4 > 0. Hence, by + radical/root rule, lim →0√ 4= √lim →0( + 4) =√4 =2 17 NegOr_Q3_Basic Calculus11_Module1_v2 3 Example 9. Evaluate lim →−2√ 2 + 3 − 6 Solution. Since the index of the radical sign in odd, we do not have to worry that the limit of the radicand is negative. Therefore, the radical/root rule implies that 3 3= 2 √ lim →−2( + 3 − 6) lim →−2√ 2 + 3 − 6 3 =√4 − 6 − 6 3 =√−8 = -2 ’s More Activity. Use the limit theorems to evaluate the following, if the limit exist. 1. lim →12 2. lim →25 3. lim →−3(4 + 2) 4. lim →32 − 4 5. lim →2(8 − 3 + 12 2) 6. lim →22 2−3 +1 3 +4 I Have Learned Generalization Directions: Reflect the learning that you gained after taking up the two lessons in this module by completing the given statements below. Do this on your activity notebook. Do not write anything on this module. What were your thoughts or ideas about the topic before taking up the lesson? I thought that _____________________________________________________________ __________________________________________________________________________. What new or additional ideas have you had after taking up this lesson? I learned that (write as many as you can) ___________________________________________________________________________ _______________________________________________________________________. How are you going to apply your learning from this lesson? I will apply ___________________________________________________________________________ _______________________________________________________________________. 18 NegOr_Q3_Basic Calculus11_Module1_v2 I Can Do Application (Performance Task) In a short sized bondpaper folded crosswise, make a booklet then write the 8 Limit Laws and give 2 examples of each. Each of the examples is worth 5 points. Examples should not be copied from the given in the previous activities. Be guided with the rubrics in scoring your output. CREATIVE PROJECT ASSESSMENT RUBRIC CATEGORY 5 4 3 2 Required Goes over Includes all Missing one Missing one Elements and above of the or more of or more of all the required the the required elements as required required elements stated in the elements as elements as stated in the directions/ins stated in the stated in the directions/ins tru ctions directions/ins directions/ins tru ctions tru ctions tru ctions Creativity Exceptional Thoughtfully A few original Shows little ly clever and uniquely touches creativity, and presented; enhance the originality unique in clever at project to and/or effort showing deep times in show some in understanding showing understandi understandin understandi ng of the g the ng of the material material material Neatness and Exceptional Attractive and Acceptably Distractingly Attractiveness ly particularly attractive but messy or attractive neat in design may be very poorly and and messy at designed. particularly lay out times and/or Does not neat in design show lack of show pride and organization in work. lay out  Mathematic No A few Several Many al mathematical mathematical mathematical mathematical Grammar mistakes in mistakes mistakes mistakes and the project which are which are throughout Solution not distracting in the project. distracting in the project Clearly not the project proofread. Source: West Mark School. Retrieved from https://www.westmarkschool.org/uploaded/photos/1617/Summer_Reading/Creative_Project_Asse ssment_Rubric.pdf 19 NegOr_Q3_Basic Calculus11_Module1_v2 Multiple Choice. Read and understand each statement. Write the letter of the correct answer on your activity sheet/notebook. 1. Which defintion below illustrates the Constant Multiple Theorem which defines the limit of a multiple of a function is simply that multiple of the limit of the function? A. lim → ( ( )) = lim → ( ( )) =. B. lim → ( ) = C. lim → ( ( ) ∙ ( )) = lim → ( ) ∙ lim → ( ) = ∙. D. lim → ∙ ( ) = ∙ lim → ( ) = ∙. 2. 3. 5. What is the limit of 2−5 +4 −1as x approaches 1 from left. A. 3 B. 1 C. -1 D. -3 6. What is the limit 2−5 +4 −1as x approaches 1 from right. A. 3 B. 1 C. -1 D. -3  20 NegOr_Q3_Basic Calculus11_Module1_v2 7. Given the table of values below. What does this show base on their limit values? x f(x) x f(x) 0.5 -3.5 1.5 -2.5 0.88 -3.12 1.17 -2.83 0.996 -3.004 1.003 -2.997 0.999 -3.0001 1.0001 -2.9999 9 A. The table shows that as the value of x increases the value of f(x), decreases. B. The table shows that as the value of x decreases the value of f(x), increases. C. The table shows that as x approached 1 from left or right, f(x) approaches -3. D. The table shows that as the value of x decreases the value of f(x) increases. 8. This symbol, lim → ( ) = is read as A. The limit of x to c as it approaches to f(x) is L. B. The limit of f(x) as x approaches to c is L. C. The limit of L as x approaches to c is f(x). D. The limit of L as f(x) approaches to c is x. 9. Evaluate3 −1= lim →13 2 9 −1 A. ∞ B. -∞ C. 2 D. 0 10. Evaluate lim →0 3−8 −4= 2 A. 4 B. 2 C. 1 D. 0 11. 12. The Power Limit states that  A. the limit of an integer power p of a function is just that power of the limit of a function. B. the limit of an integer power p of a function is just the limit of the power of a function. C. the limit of an integer power p of a function is just that power raised to another power of the limit of a function. D. the limit of a function power p of an integer is just that power of the limit of a function. 13. Which among the following shows the limit of a constant given the symbol? A. lim →1212 = 12 C. lim →64 = 64 B. lim → −3.14 = D. lim → 7 = 7 14. The limit of x as x approaches to c is equal to c. This may be thought of as the substitution law because x is simply substituted by c. lim → = as illustrated in the example below. A. lim →5 = 5 C. lim →−10−10 = −10 B. lim →0.0050.005 = D. lim 5→ = 0 1 9 2 → 3( −1 15. 3 −1). Evaluate lim A. ∞ B. -∞ C. 2 D. 0 21 NegOr_Q3_Basic Calculus11_Module1_v2 22 NegOr_Q3_Basic Calculus11_Module1_v2 23 NegOr_Q3_Basic Calculus11_Module1_v2 References Arceo, Carlene P., Lemence, Richard S. 2016. Basic Calculus Teaching Guide for Senior High School. Quezon City: Department of Education - Bureau of Learning Resources (DepEd-BLR). n.d. "https://msc.maths.nuim.ie." Mathematics Support System. https://bit.ly/3HU2wRR.  24 NegOr_Q3_Basic Calculus11_Module1_v2 For inquiries or feedback, please write or call: Department of Education – Schools Division of Negros Oriental Kagawasan, Avenue, Daro, Dumaguete City, Negros Oriental Tel #: (035) 225 2376 / 541 1117 Email Address: [email protected] Website: lrmds.depednodis.net