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§10.1 Introduction to Limits Dr .Hayk Melikyan Departmen of Mathematics and CS [email protected] H.Melikian 1 Learning Objectives for Section 10.1 Introduction to Limits The student will learn about: ■ Functions and graphs ■ Limits: a graphical approach ■ Limits: an algebraic approach ■ Limits of difference quotients H.Melikian Barnett/Ziegler/Byleen College 2 Functions and Graphs A Brief Review The graph of a function is the graph of the set of all ordered pairs that satisfy the function. As an example, the following graph and table represent the function f (x) = 2x – 1. x -2 -1 0 1 2 3 H.Melikian Barnett/Ziegler/Byleen College f (x) -5 -3 -1 1 ? ? We will use this point on the next slide. 3 Analyzing a Limit We can examine what occurs at a particular point by the limit ideas presented in the previous chapter. Using the function f (x) = 2x – 1, let’s examine what happens near x = 2 through the following chart: x 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5 f (x) 2 2.8 2.98 2.998 ? 3.002 3.02 3.2 4 We see that as x approaches 2, f (x) approaches 3. H.Melikian Barnett/Ziegler/Byleen College 4 Limits In limit notation we have lim 2 x 1 3. x2 Definition: We write lim f ( x) L 3 2 xc or as x c, then f (x) L, if the functional value of f (x) is close to the single real number L whenever x is close to, but not equal to, c (on either side of c). H.Melikian Barnett/Ziegler/Byleen College 5 One-Sided Limits We write lim f ( x) K xc and call K the limit from the left (or left-hand limit) if f (x) is close to K whenever x is close to c, but to the left lim number f ( x) Lline. of c on the real xc We write and call L the limit from the right (or right-hand limit) if f (x) is close to L whenever x is close to c, but to the right of c on the real number line. In order for a limit to exist, the limit from the left and the limit from the right must exist and be equal. H.Melikian Barnett/Ziegler/Byleen College 6 Example 1 4 2 On the other hand: 2 lim f ( x) 4 x4 4 lim f ( x) 4 x4 lim f ( x) 4 x2 lim f ( x) 2 x2 Since these two are not the same, the limit does not exist at 2. H.Melikian Since the limit from the left and the limit from the right both exist and are equal, the limit exists at 4: Barnett/Ziegler/Byleen College lim f ( x) 4 x4 7 Limit Properties Let f and g be two functions, and assume that the following two limits exist and are finite: lim f ( x) L and xc lim g ( x) M x c Then the limit of a constant is the constant. the limit of x as x approaches c is c. the limit of the sum of the functions is equal to the sum of the limits. the limit of the difference of the functions is equal to the difference of the limits. H.Melikian Barnett/Ziegler/Byleen College 8 Limit Properties (continued) H.Melikian the limit of a constant times a function is equal to the constant times the limit of the function. the limit of the product of the functions is the product of the limits of the functions. the limit of the quotient of the functions is the quotient of the limits of the functions, provided M 0. the limit of the nth root of a function is the nth root of the limit of that function. Barnett/Ziegler/Byleen College 9 Examples 2, 3 lim x 3x lim x lim 3x 4 6 2 2 x 2 2 x 2 x 2 lim 2 x 2x 8 x 4 lim x4 3x 1 lim 3x 1 13 x 4 From these examples we conclude that 1.lim f ( x) f (c) f any polynomial function 2.lim r ( x) r (c) r any rational function with a nonzero denominator at x = c x c x c H.Melikian Barnett/Ziegler/Byleen College 10 Indeterminate Forms It is important to note that there are restrictions on some of the limit properties. In particular if lim r ( x) 0 x c then finding lim xc denominator is 0. If f ( x ) may present difficulties, since the r ( x) lim f ( x ) 0 and lim g ( x ) 0 , then lim x c xc x c f ( x) g ( x) is said to be indeterminate. The term “indeterminate” is used because the limit may or may not exist. H.Melikian Barnett/Ziegler/Byleen College 11 Example 4 This example illustrates some techniques that can be useful for indeterminate forms. x2 4 ( x 2)( x 2) lim lim lim( x 2) 4 x 2 x 2 x 2 x 2 x2 Algebraic simplification is often useful when the numerator and denominator are both approaching 0. H.Melikian Barnett/Ziegler/Byleen College 12 Difference Quotients f ( a h) f ( a ) lim . Let f (x) = 3x - 1. Find h0 h H.Melikian Barnett/Ziegler/Byleen College 13 Difference Quotients f ( a h) f ( a ) lim . Let f (x) = 3x - 1. Find h0 h Solution: f (a h) 3(a h) 1 3a 3h 1 f (a) 3a 1 f (a h) f (a) 3h f ( a h) f ( a ) 3h lim lim 3. h 0 h 0 h h H.Melikian Barnett/Ziegler/Byleen College 14 Example 2 lim x2 – 3x = x2 lim x2 - lim 3x = 4 – 6 = -2. x2 x2 From this example we can see that for a polynomial function lim f (x) = f (c) x c H.Melikian 15 Indeterminate Form f ( x) lim If lim and , then f ( x ) 0 lim g ( x ) 0 xc xc xc g ( x) is said to be indeterminate. The term indeterminate is used because the limit may or may not exist. H.Melikian 16 The Limit of a Difference Quotient. f (a h ) f (a ) Let f (x) = 3x –1. Find lim h h0 f (a + h) = 3 (a + h) - 1 = 3a + 3h - 1 f ( a ) = 3 ( a ) – 1 = 3a - 1 f (a + h) – f (a) = 3h f (a h ) f (a ) 3 h lim 3 3 lim lim h0 h0 h0 h h H.Melikian 17 Practice Problems ( section 10.1) # 1, 5, 11, 13, 15, 17, 21, 39, 43, 46. 49, 51, 55, 69. Practice Problems ( section 10.2) # 1, 5, 7, 11, 15, 19, 21, 25, 27, 31, 33, 41, 53, 63. H.Melikian 18 Summary ■ We started by using a table to investigate the idea of a limit. This was an intuitive way to approach limits. ■ We saw that if the left and right limits at a point were the same, we had a limit at that point. ■ We saw that we could add, subtract, multiply, and divide limits. ■ We now have some very powerful tools for dealing with limits and can go on to our study of calculus. H.Melikian Barnett/Ziegler/Byleen College 19 WARM UP EXERCISE 3x 2x 1 f ( x) x 2 3x 2 2 Let Find ( A) xlim f ( x) 3 (B) xlim f ( x) 1 (C) lim f ( x) x2 H.Melikian 20 § 9.2 Continuity The student will learn about: continuity, continuity properties, and solving inequalities using continuity properties. H.Melikian 21 H.Melikian 22 Continuity As we have seen some graphs have holes in them, some have breaks and some have other irregularities. We wish to study each of these oddities. Then through a study of limits we will examine the instantaneous rate of change. H.Melikian 23 Definition A function f is continuous at a point x = c if lim f (x) exists 1. x c 2. f (c) exists 3. lim f (x) f (c) x c THIS IS THE DEFINITION OF CONTINUITY H.Melikian 24 Example 1 f (x) = x – 1 at x = 2. a. lim x 1 1 The limit exist! x2 b. f (2) = 1 AND ! c. lim x 1 1 f ( 2) x2 1 2 Therefore the function is continuous at x = 2. H.Melikian 25 Definition A function f is continuous on the open interval (a,b) if it is continuous at each point on the interval. If a function is not continuous it is discontinuous. H.Melikian 26 Example 2 f (x) = (x2 – 9)/(x + 3) at x = -3 x2 9 -6 a. lim x 3 x 3 b. f (-3) = 0/0 The limit exist! Is undefined! So x 9 c. lim f ( 3) x 3 x 3 2 -3 -6 Therefore the function is not continuous at H.Melikian x = -3. 27 Example 3 x a. lim x0 x f (x) = |x|/x at x = 0 and at x = 1 Does not exist! b. f (0) = 0/0 Undefined! 0 x c. lim f (0) x0 x Therefore the function is not continuous at Notice that the function is continuous at H.Melikian x = 0. x = 1. 28 Continuity Properties If two functions are continuous on the same interval, then their sum, difference, product, and quotient are continuous on the same interval except for values of x that make the denominator 0. H.Melikian 29 Some Special Functions and Continuity • A constant function is continuous for all x. • For n > 0, f (x) = xn is continuous for all x. • A polynomial function is continuous for all x. H.Melikian 30 Some Special Functions and Continuity • A rational function is continuous for all x except those values that make the denominator 0. • For n an odd positive integer, n f ( x) is continuous wherever f (x) is continuous. • For n an even positive integer, n f ( x) is continuous wherever f (x) is continuous and non negative. H.Melikian 31 Solving Inequalities Using Continuity Properties. A tool for analyzing graphs of functions is called a sign chart. We find where the function is zero or undefined to partition the number line into intervals for testing. We then test each interval to determine if the function is positive (above the x-axis) or negative (below the x-axis) in those intervals. H.Melikian 32 Constructing Sign Charts. 1. Find all partition numbers. a. Points of discontinuity – where the denominator is 0. b. Points where the function is zero – where the numerator is zero but the denominator is not. 2. Plot these partition numbers on the number line dividing the line into intervals. 3. Select a test number in each interval and determine if f (x) is positive (+) or negative (-) there. 4. Complete your sign chart showing the sign of f (x) on each open interval. H.Melikian 33 Sign Chart - Example. x 2 3 x x (x 3) f (x) 0 x2 x2 1. a. Points of discontinuity. Where the denominator is zero. x = 2 1. b. Points where f (x) = 0. Where the numerator is zero. x = 0, x = - 3 2. Place these partition values on a number line. -3 0 2 continued H.Melikian 34 Sign Chart - Example. x 2 3 x x (x 3) f (x) 0 x2 x2 -3 0 2 3. Select test numbers and determine if f (x) is positive or negative. x - 10 -1 1 10 f (x) - + - + 4. Complete the sign chart. - - - - - - - + + + + - - - - + + + + + + + -3 H.Melikian 0 2 continued 35 Sign Chart - Example. x 2 3 x x (x 3) f (x) 0 x2 x2 - - - - - - - + + + + - - - - + + + + + + + -3 0 2 Remember the plus signs mean the function is above the x-axis while the minus signs mean the function is below the x – axis. We can check this with a graphing calculator. H.Melikian 36 Summary. We have developed a definition for determining if a function is continuous. That is the function has no holes or oddities. We have developed a set of properties for limits. AND We have used sign charts to solve inequalities. H.Melikian 37 Sign Chart - Example. x 2 3 x x (x 3) f (x) 0 x2 x2 - - - - - - - + + + + - - - - + + + + + + + -3 0 2 The > 0 in the original problem means we want greater than 0, the positive signs (where the function is positive). The answer to the problem is then – 3 < x < 0 and x > 2. Or in interval notation (-3,0) and (2,). H.Melikian 38 x2 4 lim x 2 x 2 MATH 2000 Name_________________ H.Melikian Quiz #1 Date___________ 39 H.Melikian 40