Math 152 Analysis-1 PDF
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This document explores the real number system, covering topics such as rational and irrational numbers, operations with real numbers, and inequalities, including quadratic inequalities. It provides explanations and examples to help readers understand these concepts. The document likely serves as a learning resource for university-level mathematics students.
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CHAPTER 1 THE REAL NUMBER SYSTEM 1.0 Introduction The set of natural numbers (or the counting numbers) is the set N = {1, 2, 3, …}. This set is closed under + and , but not closed under – and . The inclusion of zer...
CHAPTER 1 THE REAL NUMBER SYSTEM 1.0 Introduction The set of natural numbers (or the counting numbers) is the set N = {1, 2, 3, …}. This set is closed under + and , but not closed under – and . The inclusion of zero (0) gives us the set W = {0, 1, 2, 3, …},commonly called the set of whole numbers. The set N may be extended to include all the negative whole numbers to give the set Z = {… – 3, –2 , –1, 0, 1, 2, 3, …}, which is called the set of integers. This set Z is closed under +, and –, but not closed under . A further extension of this set to include all fractions, both proper and improper, gives the set Q = { ba : a, b Z , b 0 }. This is called the set of Rational Numbers. Q is closed under all the arithmetic operations +, , –, . In the case of closure under , division by 0 must be avoided, as the result is either indeterminate (%) or infinity ( ). Note that (i) 0 Q, since o , b o is 0. b (ii) Z Q, since a1 = a, for all a Z. (iii) It is clear that N Z Q. Graphical Representation The above number systems may be represented graphically as follows: –4 –3 –2 –1 0 1 2 3 4 5 Z We may identify the subsets of Z as follows: Z– = {… –3, –2, – 1}, called the negative integers; and 1 Z+ = {1, 2, 3, …}, called the positive integers. The set {0} Z+ = {0, 1, 2, 3, …} is called the set of non-negative integers. Note that Z = Z– {0} Z+. The rational numbers may be represented by the straight line –3 –2 32 –1 12 0 1 2 1 3 2 2 5 2 3 Q Every fraction whether proper or improper may be represented by a point on this line. The Set of Irrational Numbers There are numbers that do not belong to Q, that is, they cannot be expressed in the form a , b where a, b Z. There are points on the number line which correspond to these numbers. The set of such numbers is called the set of Irrational Numbers. Examples of Irrational Numbers are: 2, 3, , e, etc. Example Prove that 2 is an irrational number. Let us assume that 2 is rational. Then 2 = a , a, b Z, b o, and such that a is in its least form. b b Then 2 2= a b2 a2 = 2b2 a2 is an even number a is even Let a = 2c, c Z. Then (2c) 2 = 2b2 i.e. 4c2 = 2b2 b2 = 2c2 2 b2 is even b is even. Since a is even and b b is even, it implies that a is not in its least form. This contradiction b is because of the false premise that 2 is rational. Therefore 2 is irrational. Exercise Prove that 3 is irrational. Note (i) Every rational number may be expressed either as a terminating decimal or as a recurring decimal. E.g. 1 = 0.5, 2 2 = 0.4 5 1 = 0.333 … or 0.3 3 1 = 0.0833 … or 0.083 12 (ii) Every irrational number can only be expressed as a non-terminating and non-recurring decimal number. E.g. 2 = 1.41423 ….. = 3.14159 ….. The Set of Real Numbers The union of the set of Rational Numbers and the set of Irrational Numbers is called the set of Real Numbers. Its graphical (or geometrical) representation is the straight line from – to + with an origin 0. 0 + Every real number corresponds to exactly one point on the line and every point on the line corresponds to a real number. Thus this line is called the Real Number Line and represents the Real Number System, denoted by R. 3 1.1 OPERATIONS WITH REAL NUMBERS Let a, b, c R, then (i) a + b R and ab R – closure law. (ii) a+b = b+a – commutative law of addition +. (iii) (a + b) + c = a + (b + c) – associative law of +. (iv) ab = b a – commutative law of . (v) (ab)c = a (bc) – associative law of . (vi) a (b + c) = ab + ac – Distributive law of over +. (vii) a + o = o +a = a – existence of an identity under +. (viii) a.1 = 1.a = 1 – existence of an identity under . (ix) a + x = x + a = o – existence of an inverse under + x is denoted by ( –a). (x) a.a–1 = a–1 a = 1 – existence of an inverse under . a –1 is denoted by 1a. Any set, such as R, which satisfies the above rules is called a field 1.2 INEQUALITIES Let a, b R. If a – b is a positive number, we say that a is greater than b or b is less than a. We write a > b or b < a. If the possibility exists that a equals b, then we write a b or b a. Theorems on Inequalities: If a, b, c R, then (i) Either a > b, a = b or a < b This is called the law of trichotomy. (ii) If a > b and b > c, then a > c This is called the law of transitivity. (iii) If a > b, then a + c > b + c. (iv) If a > b and c > o , then ac > bc. (v) If a > b and c < o, then ac < bc. 4 QUADRATIC INEQUALITIES a) If a, b R, such that ab 0 , then either a 0 and b 0 (i.e. both are positive) or a 0 and b 0 (i.e. both are negative). Example 1 Find the truth of x 2 x 2 0. Solution x 2 x 2 0 , factoring the left hand side we have ( x 1)( x 2) 0 Either (i) x 1 0 and x 2 0 x 1 and x 2 On a number line we have -1 0 -2 Figure 1.1 For which the truth set is {x: x > 2} or (ii) x 1 0 and x 2 0 x 1 and x 2 On a number line we have -1 0 -2 Figure 1.2 The truth set is {x: x > –1} From (i) and (ii) the truth set is {x : x 2} {x : x 1} Method 2 Graphical method Given x 2 x 2 0 , factoring the left hand side, we have ( x 1)( x 2) 0 Let f ( x) ( x 1)( x 2) For f ( x) 0, we have x 1 0 or x – 2 = 0. x 1 or x 2 5 Sketching f(x), we have f(x) I III -2 x -1 II Figure 1.3 The curve has divided the entire x-axis into three regions, thus region I where x 1 , II where 1 x 2 and III where x 2. For the truth set of this problem we seek the region(s) where f(x) > 0 ( x 2 x 2 0). The regions are I and III and therefore the truth set is {x : x 1} {x : x 2}. Note: If the question had been ( x 2 x 2 0) , then the solution would have been {x : x 1} {x : x 2}. Method 3 Let f(x) = x 2 x 2 ( x 1)( x 2). f(x) = 0 when x = – 1 and x = 2, so consider the figures of the factors in the following intervals. x 1 1 x 2 x2 x 1 x2 f (x) Here the solution set is where f(x) > 0, hence the truth set is {x : x 1} {x : x 2} b) If a, b R such that ab 0 , then either (i) a 0 and b 0 or (ii) b 0 and b 0 i.e. If the product of two numbers is negative, then the numbers are of opposite signs. 6 Example 2 Solve x 2 4 x 3 0. Solution For x 2 4 x 3 0, factoring the LHS, we have ( x 1)( x 3) 0 Therefore either (i) x 1 0 and x 3 0 x 1 and x 3 On a number line, we have -3 -1 0 Figure 1.4 Or x 1 0 and x 3 0 x 1 and x 3 On the number line we have -3 -1 0 Figure 1.5 3 x 1 Method 2: (Graphical method) x 2 4 x 3 0 , factoring the LHS, we have ( x 1)( x 3) 0 Let f(x) = ( x 1)( x 3). Now f(x) = 0 x + 1 = 0 or x + 3 = 0 x 1 or x 3 Sketching f(x), we have f(x) I III x -3 II -1 Figure 1.6 Here, for the solution we seek the region where f(x) < 0, which is region II, hence 3 x 1 is the solution. 7 Method 3: Let f(x) = x 2 4 x 3 ( x 1)( x 3). f(x) is zero when x 1 or x 3 , so consider the signs of the factors in the following intervals. x 3 3 x 1 x 1 x 1 x3 f (x) Here the solution set is where f(x) < 0 i.e. 3 x 1. Example 3 For what values of x is 2x2 + 5x – 3 i) negative ii) positive 1 Solution Let f(x) = 2x2 + 5x – 3 = (2x – 1)(x + 3). f(x) = 0 if x = or , x 3 , 2 So we consider the signs of the factors in the following intervals: x 3 3 x 1 2 x 1 2 2x 1 x3 f (x) i) We seek the region where f(x) < 0 i.e. 3 x 12. ii) We seek the region where f(x) > 0 i.e. x 3 x 1 2 Example 4 Find the truth set of 9 3x 2 x 2 0 Solution Let f ( x) 9 3x 2 x 2 (3 2 x)(3 x). For f ( x) 0, we have 3 x or x 3 2 Using the graphical method, we have f(x) II x I - 8 3 2 III Figure 1.7 Here we seek the region(s) of the curve where f ( x) 0. Since the coefficient of x2 is negative, the nature of the curve is like "". Example 5 Solve x 2 x. Solution Let f ( x) x 2 x x( x 1). For f(x) = 0, x = 0 or x = 1. Using the graphical method, we have; f(x) o 1 Figure 1.8 Hence, x 0 x 1 Example 6 For what ranges of values of x is 2 x 2 5x 3 positive? Solution Let y = x 2 5x 3 = 2x 1x 3 y = 0 2x –1 = 0 or x + 3 = 0 1 Thus x= or x= –3 2 1 i.e. the graph of the function crosses the x axis at x and at x 3 2 The coefficient of x 2 in the function is positive. The graph has a minimum turning point. A sketch of the graph is as shown in Figure 1.9. From the graph, the function is positive for values of x in y x 1 -3 0 2 9 Figure 1.9 1 the interval, less than 3 or greater than. Thus x 3 or x . 2 Example 1.7 For what ranges of values of x is 2 x 31 x positive? Solution Let y= 2 x 31 x y = 0 2x 3 = 0 or 1 x = 0 3 x= or x =1 2 The coefficient of x2 in the function is negative. Therefore the graph has maximum turning point. A sketch of the graph is shown. From the graph the function is positive for values of x 3 lying between 1 and. 2 3 y Thus 1 < x <. 2 3 x 1 2 2 Figure 1.11 1.3 THE ABSOLUTE VALUE OF REAL NUMBERS The absolute value of x, denoted by x is the distance of x from zero (0). That is why absolute value is never negative. Absolute value only asks ‘how far’? not in which direction. This means that x 3 , because 3 is three units to the right of zero, also 3 3 because – 3 is three units to the left of zero. Simplify each of the following (i) 0 6 6 6 (ii) 5 2 3 3 (iii) 2 5 3 3 (iv) 0(4) 0 0 (v) 2 3(4) 10 10 (vi) 4 (4) 4 (vii) (3) 2 9 9 10 Let x R. The absolute value of x, denoted by x, is defined as x, x 0 x = . (1) x, x 0 The following theorems hold for inequalities. For x, y R, (i) x y = x y (ii) x + y x + y, called the triangle inequality. (iii) x – y x y (iv) x = 0 x = 0. Note: The property (i) may be extended to n real numbers, namely x1. x2 … xn = x1 x2 … xn ; (ii) may be extended to n real numbers as x1 + x2 + … + xn x1 + x2 + … + xn (iii) For the case, x – y may be geometrically interpreted as the distance between any two points x and y on the number line. It is clear then that x – y = y – x. Worked Examples 1. Prove that (i) x y = xy (ii) x + y x + y Solution (i) We consider the four cases: x = 0, y = 0 x > 0, y > 0 x < 0, y > 0 x < 0, y < 0. If x = 0 and y = 0 , xy = 0.0 = 0 x y = 0 = 00 = x y. If x > 0 and y > 0, xy > 0, x = x , y = y and xy = xy = x y. 11 If x < 0 and y > 0, then x = x and y = y and x y < 0 xy = (xy) = (x) (y) = x y. If x < 0 and y < 0, then x = x and y = y and xy > 0 xy = xy But x y= (x) (y) = xy xy = x y. This completes the proof. (ii) We again consider the cases x=0, y=0 x > 0, y>0 x0 x < 0, y 0 and y > 0, then x + y > 0 x + y = x + y. Also x = x and y = y, so x + y = x + y. x + y = x + y. If x < 0 and y > 0; and suppose that x > y, then x + y < 0 x + y = (x + y) = x + ( y). Also x = x and y = y x + y = x + y Clearly, x + y (x + y ) = 2y < 0 x + y < x + y On the other hand if x < y, then x + y > 0 x + y = x + y Also x = x and y = y x + y = x + y. x + y (x + y ) = 2x < 0 x + y < x + y 12 Finally, if x = y, then x + y = 0. So that x + y = 0 = 0. But x + y = –x+y>0 x + y < x + y Now, if x < 0 and y < 0, then x + y < 0 x + y = – (x + y) = (– x) + (– y) = x + y Thus for all x, y , x + y x + y This completes the proof. 13 EXERCISE 1 1. Find the truth set of x 2 7 x 10 0. 2. Solve 6 x 2 x 35 0. 3. Solve (2 x 1)( x 2) 0. 4. Solve (1 2 x)( x 2) 0. 5. Solve 2 x 2 x 6 0. 6. Find the truth 35 2 x x 2 0. Note: (We can multiply both sides of this inequality by – 1) i.e. x 2 2 x 35 0 and find the truth set of x 2 2 x 35 0, which is the same as 35 2 x x 2 0). 7. Find the truth set of 3x 2 x 2 0. 8. Find the truth set of (3 2 x)(4 2 x) 0. 9. Prove that 2 is irrational. 10. Prove that 3 is irrational. 14 CHAPTER 2 POINT SETS 2.1 Definition Any set whose members are real numbers is called a one-dimensional point set. All the elements of a point set can be located on the real number line. E.g. The sets {0, 1, 2, 3, 4, 5}, {x: –1 < x < 0} are point sets. 2.2 Intervals The set of values of x such that a x b, where a, b R, is called a closed interval; it is denoted by [a, b]. The set {x: a < x < b, a, b R} is called an open interval and is denoted by (a, b). The half-open or half closed intervals are {x: a < x b} (a, b]. {x: a x < b} [a, b). The intervals indicated above are said to be bounded. There are unbounded intervals such as: {x : < x < b} (, b) or {x: x < b}, {x: a < x < } {x: x > a } (a, ) {x: < x < } {x: x R } R, or (, ). 2.3 NEIGHBOURHOODS The set of all points x such that x – a< , where a R and > 0 is called a neighbourhood of the point a x a < < x – a < a 0 (however small), at least one element x in A such that x > M – , the number M is called the least upper bound (l.u.b.) of A. Let m be a lower bound of A, then if for any > 0 (however small), at least one element x in A such that x < m + , the number m is called the greatest lower bound (g.l.b.). If x A, m , M (real) such that m x M, then the set A is said to be bounded. Examples (1) The set [0, 1] is a bounded set since for example, –2 < x < 3, x [0, 1]. The least upper bound is 1 The greatest lower bound is 0. (2) The set 1 1 1 A = 1, , , , 2 3 4 is a bounded set, since –1 < x < 2 , x A. The least upper bound is 1 and the greatest lower bound is 0. 18 2.6 THE BOLZANO -WEIERSTRASS THEOREM The Bolzano -Weierstrass theorem states that every bounded infinite set has at least one limit point. Examples (1) Show that S = { 1n : n = 1, 2, 3, … } satisfies the Weierstrass – Bolzano theorem. Determine the limit and prove it. Solution 1 1 1 S = 1, , , , . 2 3 4 S is bounded above by 2 and bounded below by –1. S is bounded. Also S is infinite set since it can be placed in 1 – 1 correspondence with a subset of itself. S has at least one limit point, by the Bolzano – Weierstrass Theorem. The limit point is 0. A deleted neighbourhood of 0 is 0 < 1n 0< , n N. 0 < 1n < n > 1 n> 1 It is clear that for every , there will always be an integer N() 1n 0 < , n > N(). 1 (2) Does the set T = { n : n = 1, 2, 3, …} satisfy the Bolzano – Weierstrass theorem? 2 Why? Does the set T have a limit point? If yes, find it and prove it. 19 Solution 1 1 1 T= , , , 2 4 8 Clearly T is bounded. For example, –1 < x < 1, x T. T is also infinite and so satisfies the Weierstrass-Bolzano theorem. Therefore, T has at least one limit point. In fact, T has only one limit point 0. To prove that 0 is a limit point of T, we show that every deleted -neighbourhood of 0 has elements of T. 1 0< 0 < ,nN 2n 1 2n 1 2n 1 n log 2 log 1 log n log 2 1 It is clear that for every , there will always be an integer N such that 0 . 2n 1 log The existence of an integer N( ) immediately greater than completes the proof. log 2 20 EXERCISE 2 1 1. Define a deleted neighbourhood of 2. 2 2. How many cluster points does the interval 1, 2 have? 1 1 1 3. Given that A 1, , , , find the 2 3 4 (i) lower bound of A. (ii) greatest lower bound of A. (iii) upper bound of A. (iv) least upper bound of A. 1 1 1 1 4. Find the cluster point(s) of 1, , 1, , 1, , 1, 2 3 4 5 1 5. If T n : n 0, 1, 2, 3 , then T is closed? 5 A. True B. False 1 6. Find the limit point(s) of an 1 1 n n 7. The set T 1, 2, 3, 4, 5 has no cluster point. A. True B. False 8. Find the cluster point(s) of n2 2 (i) A 1 (1) n 2 ; n 1, 2... 2n 1 1 (ii) B (1)n 1 ; n 1, 2... n (iii) C 1 (1)n ; n 1, 2... 9. (i) State Bolzano-Weierstrass theorem. 1 (ii) Does the set T n : n 1, 2, 3 satisfy that Bolzano-Weierstrass theorem? 3 Does the set T have a cluster point? If yes, find it and prove it. 21 CHAPTER 3 SEQUENCES 3.1 DEFINITION A set of numbers U1, U2, U3 , … in a definite order of arrangement and formed by a definite rule is called a sequence of numbers. Every number in the sequence is called a term of the sequence. If the sequence has a last term, then it is called a finite sequence. If the number of terms is infinite, then the sequence is called an infinite sequence. The nth term of the sequence will be denoted by Un, while the sequence itself will be denoted by {Un}. Examples (i) {1, 2, 3,…, 20} is a finite sequence of the counting numbers up to 20. 1 1 1 (ii) , , , is an infinite sequence with the nth term Un = 1n. 2 3 4 3.2 TYPES OF SEQUENCES (a) Arithmetic Sequence A sequence {Un} of numbers is said to be an arithmetic sequence if U n+1 – U n = d , n -------------- (1) where d is a constant. The constant d is called the common difference. The sum of the first n terms of an Arithmetic sequence is given by n Sn 2a (n 1)d , --------------- (2) 2 where a is the first term of the sequence. E.g. The sequence 2, 5, 8, 11, … 22 is an arithmetic sequence with a 2 and d 3. The sum of the first 10 terms is: S10 = 10 {2 (2) + (10 – 1) (3) } 2 = 5 {4 + 27} = 5 {31} = 155 (b) Geometric Sequence A sequence {Un} is said to be a geometric sequence if U n1 r, n ------------------(3) Un where r is a constant. The number r is called the common ratio. The sum of the first n terms of a geometric sequence is given by a(1 r n ) Sn , for r 1 --------------- (4a) 1 r a(r n 1) and Sn , for r 1 ---------------- (4b) r 1 where a is the first term. E.g. The sequence 1 1 1 1, , , , 2 4 8 1 is a geometric sequence with a 1 and r . 2 SUM TO INFINITY The sum to infinity of a geometric sequence dented by a S ---------------- (4c) 1 r 23 (c) Sequences which are neither arithmetic nor geometric There are sequences which are neither arithmetic nor geometric. In what follows general properties of sequences will be discussed. 3.3 LIMIT OF A SEQUENCE An infinite sequence {U n} n 1 is said to have a limit l as n approaches if Un can be made as close to l as we please by choosing n sufficiently large. We then write. lim Un = l. ------------------ (5) n The above definition may be more rigorously written as follows: lim Un = l if for every > 0 n (however small), a positive integer N (depending on ) s.t. Un l < , n > N. Examples 1 1 (1) The sequence has limit 0. Thus Un = and l = 0. n n 1 n 1 The difference between Un and l is 0 n 1 n 1 1 , n n 1 Thus any choice of n greater than would satisfy the required condition. 1 It is clear that no matter how small the difference 0 is made there will always be n the number N. 24 n 1 (2) The sequence has limit 1. n n 2 1 (3) The sequence has no limit. n Note: If a sequence has a limit then it is called a convergent sequence; otherwise it is called a divergent sequence. 3.4 THEOREMS ON SEQUENCES Theorem 1 If a sequence {Un} has a limit, then the limit is unique. (In other words, a convergent sequence has only one limit). Proof: Suppose that {Un} has two limits l1 and l2. Then > 0, N ( a + ve integer) such that Un – l1 < ɛ, n > N, and Un – l2 < ɛ , n > N. Then l1 – l2 = l1 – Un + Un – l2 l1 – Un + Un – l2 < ɛ + ɛ i.e. l1 – l2< 2 ɛ Since > 0 (however small), it follows that l1 – l2 is less than every positive integer. l1 – l2 = 0 l1 – l2 = 0 l1 = l2. This completes the proof. 25 Theorem 2 Let U n and Vn be two sequences such that lim U n A and lim Vn B , Then the following n n results hold: (1) lim Un Vn lim Un lim Vn n n n = A + B. (2) lim Un Vn lim Un lim Vn n n n = A – B (3) lim Un Vn lim Un lim Vn n n n = A B U lim U n n A (4) lim n provided that lim Vn = B 0. n V n n nlim Vn B Un (a) If B = 0 and A 0, then lim does not exist. n Vn Un (b) If B = 0 and A = 0, then lim may or may not exist. n Vn p (5) lim U np lim U n = Ap n n lim u n n (6) lim pun p = pA. n For the conditions (5) and (6), p must be real and Ap or pA must exist. 26 3.5 BOUNDED, MONOTONIC SEQUENCES A sequence {Un} is said to be bounded above if M (a constant) such that Un M, n. M is called an upper bound. If m a constant such that Un m , n , then the sequence {Un} is said to be bounded below and m is called a lower bound. If m, M such that m Un M, n, then the sequence {Un}is said to be a bounded sequence. An upper bound M is called the least upper bound (l. u. b.) of the sequence if for every ɛ > 0, there will always exist at least one term of the sequence greater than M – ɛ. A lower bound m is called the greatest lower bound (g. l. b.) of the sequence if for every ɛ > 0, at least one term of the sequence is less than m + ɛ. Example 2n 1 1 The sequence 2 is bounded below by 1 and bounded above by 4. Thus n n 2n 1 is a bounded sequence. n The least upper bound is 3 and the greatest lower bound is 2. A sequence {Un} is said to be a monotonic increasing sequence or a non-decreasing sequence if Un Un+1 , n. If Un+1 Un, n, then the sequence is called a monotonic decreasing sequence or a non-increasing sequence. If Un < Un+1, n, the sequence is said to be strictly increasing. If Un+1 < Un, n, the sequence is strictly decreasing. If a sequence is either monotonic increasing or monotonic decreasing, it is said to be a monotonic sequence. 27 Example n 1 Show that the sequence is: n (i) a monotonic decreasing sequence (ii) a bounded sequence. Solution n 1 (i) Un = n n 2 n 1 Un+1 – Un = n 1 n n n 2 n 1 n 1 n n 1 n 2 2 n n 2 2n 1 n n 1 1 = 0 n n 1 Un+1 – Un 0 Un+1 Un, n. {Un} is a monotonic decreasing sequence. n 1 2 3 4 5 (ii) , , , , n 1 2 3 4 The sequence is bounded above by 3 (say) and bounded below by 0 (say). The sequence is bounded. 28 Theorem: Every bounded monotonic sequence has a limit. Example n Use the above theorem to establish that the sequence has a limit. n 1 n 1 n Prove that lim = 1. n n 1 Solution n 1 2 3 The sequence , , , is bounded above by 2(say) and bounded below n 1 2 3 4 by 0 (say). the sequence is bounded. The sequence is monotonic increasing since n 1 n U n 1 U n n2 n 1 n 1 n n 2 2 n 1 n 2 1 = 0, n n 1 n 2 Un+1 – Un 0 Un+1 Un , n. Since the sequence is bounded and monotonic, it has a limit. n To prove that lim = 1, we wish to show that for every > 0, a positive integer N, n n 1 n such that 1 < n > N. n 1 n n n 1 1 1 Now 1 = = =. n 1 n 1 n 1 n 1 29 For any > 0 1 < n 1 (n + 1) > 1 1 n> 1. 1 Thus choosing N a positive integer immediately larger than ( 1 ) , the result follows. Example Prove that 4 2n n 2 3 (a) lim (b) lim = 0 n 3n 2 n 4 3 Solution (a) For every > 0 4 2n 2 < 3n 2 3 4 2n 2 < 3n 2 3 3 4 2n 2 3n 2 < 3 3n 2 16 16 < < 3 3n 2 3 3n 2 16 16 (3n + 2) > 3n + 2 > 3 3 1 16 n> 2. 3 3 1 16 Hence choosing N immediately larger than 2 , the result follows. 3 3 30 (b) For every > 0 , n 3 0 < 4 n 3 < 4 3 n loge < loge 4 log e 3 n> (since loge f(x2), then f(x) is strictly decreasing. Examples (1) The function f ( x) 2 x 1 is monotonic increasing in the interval [0, 1]. 1 (2) The function f ( x) is monotonic decreasing in the interval [1, 3]. x (3) The function x 1 f ( x) x is monotonic decreasing in the interval [1, 5]. MAXIMA AND MINIMA Let x0 [a, b]. If a function f(x) is such that f(x) f(x0), x [a, b], then f(x) is said to have an absolute maximum in the interval [a, b] at x = x0. The maximum value is f(x0). If f(x) f(x0), x [a, b], then f(x) is said to have an absolute minimum in the interval [a, b] at x = x0. The minimum value is f(x0). If the above statements are true only for values of x in some deleted - neighbourhood of x0 then f(x) is said to have a relative maximum or a relative minimum at x0. 4.6 TYPES OF FUNCTIONS There are three major types of functions: Polynomial functions, algebraic functions and transcendental functions. 1. Polynomial Functions These are functions of the form Pn(x) = a0 + a1 x + a2 x2 + + an xn , an 0. ------------------ (1) This is called a polynomial in x of degree n. The numbers a0, a1, …, an are constants and n is a positive integer. Theorem Every polynomial equation f ( x) 0 has at least one root. If the degree of the polynomial is n , then the equation Pn(x) = 0 has exactly n roots. 35 2. Algebraic Function Any function y = f(x) satisfying an equation of the form P0 (x) yn + P1 (x) yn1 + + Pn-1 (x)y + Pn (x) = 0 , ---------------- (2) where P0 (x), , Pn (x) are polynomials in x , is called an algebraic function. If an algebraic function may be expressed as the quotient of two polynomials, i.e. P(x)/Q(x), where P(x) and Q(x) are polynomials, then the function is called a rational algebraic function. Otherwise it is an irrational algebraic function. 3. Transcendental Functions Any function which is not algebraic is called a Transcendental Function. A transcendental function is expressible as an infinite sum of polynomials. Examples (a) 2x4 + x3 – 2x2 + x + 9 ------------------ a polynomial of degree 4. 1 (b) y = 3 is an algebraic function since x 1 x3. 3 – 1= 0 (x3, – 1 are polynomials) x x (c) y = is algebraic, since x 1 x x x. + – x = 0 x 1 x 1 x (x + 1) – x = 0 x 1 4.7 ELEMENTARY TRANSCENDENTAL FUNCTIONS 1. Exponential Functions. These are functions of the form f(x) = ax , a 0, 1 2. Logarithmic Functions These have the form f(x) = loga x, a 0, 1 If a = 2.71828 … = e , we write f(x) = logex lnx usually called logarithm to the natural base. 3. Trigonometric Functions These are the circular functions sinx, cos x and tanx. 36 There are also their reciprocals 1 1 1 sec x = , cosec x = and cot x =. cos x sin x tan x sin x cos x Furthermore, tan x = and so cot x =. cos x sin x The following results hold: (a) sin2x + cos2x = 1 – Pythagoras’ theorem (b) 1 + tan2x = sec2x (a) 1 + cot2x = cosec2 x (b) sin (x + y) = sin x cos y + cos x sin y (c) cos (x + y) = cos x cos y – sin x sin y tan x tan y (d) tan (x + y) = 1 tan x tan y 4. Inverse Trigonometric Functions These are the functions f(x) = sin–1x (also called arc sine) f(x) = cos–1x (arc cosine). f(x) = tan–1 x (arc tangent) f(x) = cosec–1x (arc cosech x) f(x) = sec–1 x (arc secant x) f(x) = cot–1x (arc cotangent x) 5. Hyperbolic Functions These are functions defined in terms of the exponential functions as follows: 1 x –x (a) sinh (x) or sinh x = (e – e ). 2 1 x (b) coshx = (e + e–x ). 2 sinh x e x e x (c) tanh x = x cosh x e e x 1 2 (d) cosech x = x sinh x e e x 1 2 (e) sech x = x cosh x e e x 1 e x e x (f) coth x = x tanh x e e x 37 6. Inverse Hyperbolic Functions These include the following: (a) sinh–1 x (b) cosh–1 x (c) tanh–1 x (d) cosech–1 x (e) sech–1 x (f) coth–1 x. (See notes on hyperbolic functions). GRAPHS OF FUNCTIONS Let us now consider the graphs of certain special functions. (1) Exponential Functions f(x) = ax, a 0, 1 , x R. For a > 1 , When x = 0, a0 = 1 As x , ax 0+ As x + , ax f(x) = a–x , a > 1 , x R. When x = 0, a0 = 1 As x , ax + As x + , ax 0+ The two graphs are as shown below on the same axes. 38 y y ax y a x (0,1) x 0 Figure 4.7.1 The exponential function defined by f(x) = ex, where e 2.72 is a special exponential function. The number e is called the natural base. How the number e was arrived at will be discussed later. (2) Logarithmic Functions f(x) = loga x , a > 1 , x R , When x = a, f(a) = 1 When x = 1, f(1) = 0. As x 0+, f(x) . x cannot assume – ve real numbers. Loga x is also the inverse function of the exponential function y = ax. The graph is shown below 39 y a 1 y log a x a x 1, 0 a 1 a 1 Figure 4.7.2 When a = 10, then log10 x is called the common logarithm. Relationship with the exponential function y = ax , a > 1. The functions y = ax and y = loga x are mutually inverse functions. The two graphs are represented on the same graph. y y ax yx y log a x 0,1 x 1, 0 Figure 4.7.3 40 4.8 EVEN, ODD AND PERIODIC FUNCTIONS A function f(x) is said to be an even function if f(– x) = f(x) , x ------------------- (1) A function is called odd if f(– x) = – f(x) , x ------------------- (2) You may be asked to "determine algebraically" whether a function is even or odd. To do this, you take the function and substitute –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f(–x) = f(x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f(– x) = –f(x), so all of the "plus" signs become "minus" signs, and vice versa), then the function is odd. In all other cases, the function is "neither even nor odd". Example 1 Determine algebraically whether f(x) = –3x2 + 4 is even, odd, or neither. f(–x) = –3(–x)2 + 4 = –3(x2) + 4 = –3x2 + 4 =f(x) Which means that f(x) is even. Example 2 Determine algebraically whether f(x) = 2x3 – 4x is even, odd, or neither. Solution f(–x) = 2(–x)3 – 4(–x) = 2(–x3) + 4x = –2x3 + 4x The final expression is the exact opposite of the function I started with, by which I mean that the sign on each term has been changed to its opposite, just as if I'd multiplied through by –1: –f(x) = –1[f(x)] = –[2x3 – 4x] = –2x3 + 4x This means that f(x) is odd. 41 Example 3 Determine algebraically whether f(x) = 2x3 – 3x2 – 4x+4 is even, odd, or neither. Solution f(–x) = 2(–x)3 – 3(–x)2 – 4(–x) + 4 = 2(–x3) – 3(x2) + 4x + 4 = –2x3 –3x2 + 4x + 4 This is neither the same function I started with (namely, 2x3 – 3x2 – 4x + 4) nor the exact opposite of what I started with (namely, –2x3 + 3x2 + 4x – 4). This means that f(x) is neither even nor odd. Remarks 1) There are functions which are neither even nor odd, and many not periodic. E.g. f(x) = x2 + x3 2) Every function f(x) may be expressed as the sum of an even and an odd function, since f(x) = 1 {f(x) + f(– x) } + 1 {f(x) – f(– x)}. 2 2 Even odd. E.g. For f(x) = (x + 1)2. f(– x ) = (– x + 1)2 f(x) + f(– x) = (x + 1)2 + (1 – x )2 is an even function; while f(x) – f(– x) = (x + 1)2 – (1 – x )2 is an odd function. Special Even and odd functions (a) f(x) = sin x is an odd function since sin (– x) = – sin x. (b) f(x) = cos x is an even function since, cos (– x) = cos x. Symmetry An even function is symmetrical about the y-axis. An odd function has a rotational symmetry of order 2 about the origin. Examples (1) f(x) = x2 is an even function f(– x) = (–x )2 = x2 = f(x) 42 y y x2 x 0 (2) f(x) = x3 is an odd function: f(– x) = (–x) 3 = (–1) 3 x3 = –x 3 = –f(x) y y x3 x A function f(x) is said to be a periodic function if a real number T such that f(x + T) = f(x) , x ------------------- (3) (3) (a) f(x) = sin x is periodic with period 2 f(x + 2) = sin (x + 2) = sin x = f(x) 43 y 1 y sin x x 2 3 4 -1 (b) f(x) = tan x is periodic with period f(x + ) = tan (x + ) = tan x = f(x). y x 3 5 2 2 2 3 Note: A periodic function repeats its value after every value x = T. 44 EXERCISE 4 (1) Consider the function x f ( x) x 1 in the interval [0, 1]. Investigate the following properties of the function f. (a) Boundedness. (b) Monotonicity. (2) Consider the function 1 f ( x) x x in the interval [1, 3]. Describe the following properties of the function: (e) Boundedness (f) Monotonicity (3) Show that (a) f ( x) x sin x is an even function (b) f ( x) x cos x is an odd function. (4) From the function f ( x) ( x 1)2 , construct (a) an even function (b) an odd function 45 CHAPTER FIVE LIMITS 5.1 LIMITS OF FUNCTIONS 5.1.1 Definition Let f(x) be a function defined in a deleted - neighbourhood of the point x = x0. The function f(x) is said to have the limit l as x approaches x0 if f(x) can be made as close to l as we please by choosing x sufficiently close to x0. We write lim f(x) = l. ------------------- (1) x x0 In a rigorous language we say that lim f(x) = l if for every > 0 (however small, > 0 x x0 ( = () ) such that f(x) l < , whenever 0 < x x0 < . 5.2 Right Hand /Left Hand Limits The lim f(x) as x x0 usually depends on how x approaches x0. Since x0 is a real number, x can approach x0 in only two possible directions. (See the figure). x x 0 x x 0 x0 Different results could be obtained for each of the cases. Let us consider the following examples: 1) lim (2x +1) = 1 x 0 lim (2x+ 1) = 1 x 0 1 2) lim = x0 x 1 lim = + x0 x 1 3) lim tan 1 = x0 x 2 1 lim tan 1 = x0 x 2 46 4) Let f(x) be defined by 1, x 0 f (x) = 1, x 0 (See Fig. 1) +1 0 1 Fig. 1 lim f (x) = 1 x 0 lim f (x) = +1 x 0 The above examples illustrate the fact that lim f (x) lim f(x) (in general). x x0 x x0 Let lim f (x) = l1 , and x x0 lim f (x) = l2. x x0 Then l1 is called the left hand limit, while l2 is called the right hand limit of f(x) as x approaches x 0. In general l1 l2. However, if l1 = l2 = l then lim f(x) = l. ------------------- (2) x x0 47 (3) Define x 1 , x 1 x 1 f(x) = 0, x 1 (i) Draw the graph (ii) (ii) show that lim f(x) does not exist. x 1 Solution Graph : for x < 1 , x 1 = – (x 1) f(x) = x 1 = – 1 x 1 for x > 1, x 1 = x 1 x 1 f(x) = = 1 x 1 f(1) = 0. y f(x) = 1 x 1 f(x) = – 1 Fig. 2 x 1 (ii) lim f (x) = lim = lim (– 1) = – 1 x 1 x 1 x 1 x 1 x 1 lim f (x) = lim = lim (+ 1) = 1 x 1 x 1 x 1 x 1 lim f (x) lim f (x) x 1 x 1 48 lim f (x) does not exist. x 1 5.3 THEOREMS ON LIMITS Let lim f(x) = A, lim g(x) = B , where A and B are finite constants. x x0 x x0 Then (a) lim (f (x) + g(x) ) = lim f(x) + lim g(x) x x0 x x0 x x0 = A + B (b) lim (f (x) g(x) ) = lim f(x) lim g(x) x x0 x x0 x x0 = A B (c) lim (f (x). g(x) ) = lim f(x). lim g(x) x x0 x x0 x x0 = A. B lim f x f x x x0 A (d) lim . x x0 g(x) lim gx x x0 B provided that B o. f x (i) If A o and B = o, lim does not exist. x x0 g(x) (ii) If A = 0 and B = 0, the form 0 is indeterminate and the limit may or 0 may not exist. 5.4 Evaluation of Limits The following rules will assist us when evaluating limits of functions (1) If f(x) is a polynomial in x, then lim f (x) = f(x0). x x0 E.g. lim (x2 + 2x – 5) = 12 + 2 (1) – 5 = – 2. x 1 (2) If f(x) is an elementary function, then lim f (x) = f(x0). x x0 49 E.g. lim sin x = sin 0 = 0. x 0 (3) If f(x) = x x0 gx , x x0 and is real, then x x0 hx g(x) lim f(x) = lim. x x0 x x0 h(x) x3 8 E.g. For f(x) = , x 2 x 2 lim f(x) = lim x3 8 = lim x - 2 x 2 2 x 4 x 2 x 2 x 2 x 2 x - 2 = lim (x2 + 2x + 4) x 2 = 22 + 2(2) + 4 = 12 x 1 (2) lim 1 = e x x 1 x x 1 (3) lim = e x 0 5.5 Limit as x and as x If f (x) approaches a number L as x →∞, then we write lim L. x In similar way, we write lim L if f (x) approaches a number L as x →-∞. x Examples 1 1. Let f ( x) , x 0. x As x , or as x , f (x) takes on values closer and closer to zero. 1 1 Thus lim 0 = lim. x x x x 3x 2 7 x 6 2. Evaluate lim . x 4 x 2 3 x 6 50 Here, we divide each term by the height degree of the numerator and the denominator, i.e., 2 3x 2 7 x 6 7 6 3x 7 x 6 2 2 2 2 3 x 3 x x x x2 lim 2 x 4 x 3 x 6 = x 2 lim = lim 4 3 6 x 4 x 3x 6 4 x 2 x2 x2 x x2 The greatest integer function truncation or “birthday’’ function, indicated by [ x ], is the function that drops any fractional quantity over an integer. For example, [2.99999] = 2 and [ ] = 3. As an example lim [ x] 2 , but lim [ x] 2. x 2 x 2 5.6 L’ HOSPITAL’S RULE f ( x) Suppose we have to find the limiting value of a function F ( x) at x c , when direct substituting of g(x,) g ( x) 0 x c gives the indeterminate form i.e., at x c , f ( x) 0 and g ( x) 0. 0 In this case, the ratio of the differential coefficients of the numerator and denominator at x = c provided, of course, that both f (c) and g (c) are not zero themselves. f ( x) f ( x) :. lim lim x c g ( x) x c g ( x) This is known as L’ HOSPITAL’S rule and is extremely useful for finding limiting values when the differential coefficients of the numerator and denominator can easily be found. Example x 3 x 2 x 1 Find lim 2 x 1 x 2x 3 Solution x 3 x 2 x 1 0 lim 2 = (Indeterminate) x 1 x 2x 3 0 3 x 2 2 x 1 4 lim =1 x 1 2x 2 4 x 3 x 2 x 1 :. lim 2 =1 x 1 x 2x 3 51 Example x 2 sin 3x Find lim 2 x 0 x 4x Solution x 2 sin 3x 0 lim 2 Direct substitution gives , so we apply L’ HOSPITAL’S rule which gives x 0 x 4x 0 2 x 3 cos 3x 3 lim x 0 2x 4 4 Example x sin x Determine lim x x 0 2 Solution x sin x 0 lim (indeterminate ) x 0 x 0 2 Applying L’HOSPITAL’S rule, we have sin x 0 lim 0 x 0 2x 2 5.7 Other Indeterminate Forms Other indeterminate forms include , ∞ - ∞, and 0. Examples 1 1 1. Evaluate lim x 0 x sin x This expression is an example of an indeterminate form of the type ∞ - ∞ as x → 0. In order to use L’ 1 1 sin x x HOSPITAL’S rule, we first rewrite the expression as , and this latter expression is of the x sin x x sin x 0 form as x→ 0. 0 52 1 1 sin x x Using L’ HOSPITAL’S rule, we obtain lim = lim x 0 x sin x x 0 x sin x cos x 1 0 = lim = (Indeterminate) x 0 sin x x cos x 0 sin x Applying L’ Hospital’s rule again, we have lim = 0 x 0 2 cos x x sin x 1 1 lim = 0 x 0 x sin x 2. Evaluate lim x ln x. This is of form 0. ∞. x 0 0 To use L’Hospital’s rule we must first write xlnx so that it has the form or as x →0+. 0 ln x xln x = 1 x Where both numerator and denominator of the right side become infinite as x → 0+. Using L’Hospital’s rule ,we obtain 1 ln x lim x ln x = lim lim x = lim { x} 0. x 0 x 0 1 x 0 21 x 0 x x 53 EXERCISE 5 (1) A function f(x) is defined as x3 1 f(x) = , x 1. x 1 Evaluate lim f(x). x 1 (2) Draw the graph of the function f(x) = x 2 By determining lim f(x) and lim f(x) show that lim f(x) exists. x2 x 2 x2 (3) Evaluate the following: (a) lim x 2 x 2 3 x 2 x2 4 (b) lim exsin2x x0 1 1 2 (c) lim x 1 x 1 x 3 3x 5 (4) A function f(x) is defined by 1, x 1 f(x) = x 1 x 1 1, x 1 (i) Draw the graph of f(x). (ii) Show that lim f(x) and lim f(x) both exist. x 1 x 1 54 1. Determine: 1 2 1 1 2 2 a) lim b) lim 2 c) lim 5 x 0 x x x 0 x x x 0 x x 1 x 3x 2 4 x 5 x3 2. a)lim b) lim 3 c) lim x 1 x x 2 x 6 x 2 x 1 x x 3 3 In each of problems 3 through 7, determine whether lim f ( x) , lim f ( x) , and lim f ( x) exist. x c x c x c Evaluate each limit that does exist. 1 x2 1 x 1 3. f ( x) C 1 x, x 1 1 x2 , x0 4. f ( x) C0 x2 , x0 sec x, x0 2 5. f ( x) C 2 cos ecx , 0 x 2 sin x x 2 0, x C 6. 2 2 sin 3x, x 2 sin x, x0 7. f ( x) 1, x0 C0 sin 3x, x0 55 L’HOSPITAL’S RULE Find the following: sin x 1 cos x i.) lim ii) lim x 0 x x 0 x2 ln x x sin x iii) lim iv) lim 2 x 1 x2 1 x 0 x x ln x sin x x v) lim vi) lim x 0 cot x x 0 tan x x 1 cos 2 x vii) lim viii) lim sec x tan x x 0 2 x x 2 1 sin 2 x ix) lim ln x x) lim 2 0 x x 0 2x 56 CHAPTER 6 CONTINUITY AND DIFFERENTIATION 6.1 CONTINUITY OF A FUNCTION Definition: A function f ( x) is said to be continuous at a point x x 0 if the following conditions hold: (i) lim f ( x) exists, x x0 (ii) f ( x0 ) is defined, and (iii) lim f ( x) f ( x0 ) x x0 Any function which fails any one of the above three conditions at x0 is said to be discontinuous at x0. Examples x2 8 (i) Given that f ( x) . Is f continuous at x 3? x3 Solution f is not defined at x=3. Hence f is not continous at x=3. 1 , if x 0 (ii) Given that g ( x) x. Is g continious at x 0? 2, if x 0 Solution g fails to be continuous at x = 0, since lim g ( x) does not exist. x 0 x2 9 , if x 3 (iii) Given that h( x) x 3. Is h continious at x 3? 7, if x 3 57 Solution lim h( x) 6 and h(3) 7, therefore lim h( x) h(3). Hence h is not continious at x 3 x 3 x 3 x2 9 , if x 3 (iv) Given that h( x) x 3. Is h continious at x 3? 6, if x 3 Solution lim h( x) 6 and h(3) 6, therefore lim h( x) h(3). Hence h is continious at x 3 x 3 x 3 Equivalently we could say that f ( x) is continuous at the point x x0 if f ( x0 ) is defined and lim f ( x) f ( x0 ) x x0 Using the definition we say that f(x) is continuous at x = x 0 if given > o (however small) > o such that f(x) f(x0) < , whenever x x0 < . Right Hand/Left Hand Continuity If only lim f(x) = f(x0), where f(x0) is defined, then f(x) is continuous on the left; if only lim x x0 x x0 f(x) = f(x0), then f(x) is continuous on the right. Example x2 , if x 3 Given that f ( x) . Is f continious at x 3? 3, if x 3 Solution lim f ( x) 3 x 3 lim f ( x) 3 x 3 lim f ( x) lim f ( x). Therefore lim f ( x) doesnot exist. x 3 x 3 x 3 Hence f is not continious at x 3 Continuity in an Interval A function f(x) is said to be continuous in an interval (a, b) if f(x) is defined for every x0 (a, b) and lim f (x) = f(x0). x x0 For the closed interval [a, b], f(x) is continuous in [a, b] if f(x) is continuous in (a, b) and in addition 58 lim f (x) = f(a) and lim f (x) = f(b), where f(a) and f(b) are defined. x a x b THEOREMS ON CONTINUITY 1. If f(x) and g(x) are both continuous at a point x = x0, then f(x) g(x), f x f(x). g(x) and are all continuous at x = x0 (provided that g(x0) 0 in the last g x case). 2. Every polynomial in x, Pn(x) is continuous for all real x. 3. If y = f(x) is continuous at x = x 0 and z = g(y) is continuous at y = y0 and if y0 = f(x0), then the function z = g{f(x) }, called the composite or function of a function, is continuous at x = x0. That is, a continuous function of a continuous function is continuous. 4. If f(x) is continuous in a closed interval, then it is bounded in the interval. 6.7 DIFFERENTIABILITY Let f(x) be defined at some point x0 in the interval (a, b). Then, the derivative of f(x) at x0 denoted by f (x0) is defined as f x f x 0 f (x0) = lim , if the rhs limit exists. x x0 x x0 Setting h = x – x0, the above definition becomes f x 0 h f x 0 f (x0) = lim ------------------- (4) ho h A function f(x) is said to be differentiable at a point x0 if f (x0) exists. Right Hand/Left Hand Differentiability A function f(x) is said to be left hand differentiable at x = x0 if f x 0 h f x 0 f ' (x0) = lim exists. -------------------- (5) ho h It is right hand differentiable at x = x0 if 59 f x 0 h f x 0 f ' (x0) = lim exists. ----------------------(6) ho h The function f(x) is said to be differentiable at x = x0 if and only if f ' (x0) = f ' (x0). Differentiability in an Interval A function f(x) is differentiable in an interval (a, b) if f (x0) exists x0 in (a, b). In particular, for the closed interval [a, b] f (x0) must exist x0 (a, b) and f ' (a) and f ' (b) must exist. Examples (1) Show that the function f(x) = x2 is differentiable at x = 2. Solution f 2 h f 2 f (2) = lim h 0 h i.e. 2 h 2 2 2 f (2) = lim h0 h 4 4h h 2 4 4h h 2 = lim lim h 0 h 0 h h = lim {4 h} h 0 = 4. f (2) exists and so f(x) is differentiable at x = 2. (2) Let x2 , x 1 f(x) = 1 , x 1 x , x 1 Find (a) f ' (1) (b) f ' (1) Is f(x) differentiable at x = 1? 60 Solution f 1 h f 1 1 h 1 f ' (1) = lim = lim ho h ho h h = lim = lim 1 = 1 ho h h o f ' (1) = lim f 1 h f 1 = lim 1 h 2 1 ho h ho h 2h h 2 = lim h o h = lim (2 + h) = 2 h o Clearly then, f ' (1) f ' (1) f (1) does not exist, and so f(x) is not differentiable at x = 1. THEOREM If a function f(x) is differentiable at a point x = x 0, then it is continuous there. The converse, however, is not true. Proof: f ( x h) f x0 f(x0 + h) – f(x0) = 0 . h h f ( x0 h) f x0 lim f x0 h f x0 lim . lim h h0 h0 h h 0 = f (x0). lim h h0 = 0 , since f (x0) exists. lim f(x0 + h) = f(x0) h0 Setting h = x – x0 , we have lim f(x) = f(x0) x x0 which completes the proof. 61 The converse of this theorem is false. Consider the counter example x2 , x0 f(x) = x , x0 y 0 , x0 y = x2 lim f(x) = lim x = 0 x 0 x 0 lim f(x) = lim x2 = 0 o x x 0 x 0 f(o) = 0 y=x f(x) is continuous at x = 0. Graph of y = f(x). Now f (0 h) f 0 f h f 0 f (o) = lim lim h0 h h 0 h h 0 = lim h 0 h = 1 f ( h) f 0 f + (0) = lim h0 h h2 0 = lim lim h = 0 h 0 h n 0 f (0) f + (0) f(x) is not differentiable at x = 0. 62 6.8 DIFFERENTIATION The derivative of a function f(x) at any point x is defined by f x h f x f (x) = lim . h0 h The function f (x) is also called the Derived Function of f(x). The process of finding the derivative is called differentiation. The derived function f (x) enables one to obtain the derivative of f(x) at any point x = x0. The following cases will be considered. 1. f(x) = xn , n rational. (i) for f(x) = x f x h f x f (x) = lim . h0 h ( x h) x h = lim lim = 1 ------------- (7) h 0 h h 0 h (ii) f(x) = x2 x h 2 x 2 2hx h2 f (x) = lim h0 hlim 0 . h h = lim {2 x h} = 2x ----- (8) h 0 (iii) f(x) = x3 x h x 3 3 f (x) = lim h0 h 3x 2 h 3xh2 h3 lim h 0 h lim {3x 2 3xh h2 } = 3x2 --------------- (9) h 0 63 (iv) f(x) = xn , n integer. x h x n n f (x) = lim h0 h n n n1 n n2 2 n n1 x x h x h xh hn xn lim 1 2 n 1 = h0 h n n n n2 = lim x n1 x n2 h xh hn1 h 0 1 2 n 1 n n 1 n! = x x n 1 n x n 1 ------------- (10) 1 n 1 ! 1! 1 (v) f(x) = x1 = x 1 1 x h x f (x) = lim h0 h x x h h = lim lim h0 h x x h h 0 h x 2 xh 1 1 = lim 2 2. ----- (11) h0 x xh x The same result as for (iv) holds for negative integers. In fact for all rational numbers n. f(x) = xn f (x) = n xn–1 64 2. Trigonometric Functions. (i) f(x) = sin x. sin x h sin x f (x) = lim h0 h A B A B Using the fact that sin A – Sin B = 2 Cos. Sin we have 2 2 x h x x h x 2 cos sin f (x) = lim 2 2 h0 . h 2x h h 2 cos sin = lim 2 2 h0 h h sin h 2 = lim cos x lim = cos x. 1 h0 2 h0 h 2 = cosx ------------- (12) (ii) f(x) = cos x f x h f x f (x) = lim h0 h cos x h cos x = lim h0 h x h x x h x 2sin sin 2
