Good Day, Franciscans! Academics. And beyond. Introduction to the Limits Functions Unit I – Module 2 Grade 11 | Precalculus CALCULUS The study of the RATE OF CHANGE. Calculus is also called an INFINITESIMAL Calculus. CALCULUS Primarily deals with the limit process as it quantifies the relationship between two variables or quantities Precalculus Limit process Calculus Limit of a Function Let f be a function defined at every number in some open interval containing c, except possibly at the number c itself. If the value of f is arbitrarily close to the number L for all the values of x sufficiently close to c, then the limit of f(x) as x approaches c is L. lim π π = π³ π→π Limit and Function value The limit of a function as it approaches c is not necessarily equal to its value c. Thus, lim π(π) π→−π can assume a value different from f(c) Existence of Limit The limit of a function as x c exists if - f(c) is defined; or - f approaches the same value as x moves closer to c from both directions. Numerical representation of the Limits of Functions The limit of function refers to the value that a “function” approaches a specific value. π₯−4 Consider the function defined by f x = 2 π₯ −16 0−4 f 0 = 2 0 − 16 1 f 0 = 4 As the values of x come close to 0, then the values of f(x) come close to 1 or 0.25. 4 Table of values for f x π₯−4 = 2 as x approaches 0 π₯ −16 x approaches 0 from the left x approaches 0 from the right x -0.1 -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01 0.1 f(x) 0.2564 0.2506 0.2501 0.2500 0.25 0.24999 0.24993 0.24937 0.24390 f(x) approaches 0.25 f(x) approaches 0.25 Your Turn! 3 Evaluate lim (−π₯ +13π₯2 − 56π₯ + 83) π₯→4 The limits law for polynomial, Rational, and radical function Limit Laws Let c and k be real numbers so that lim π(π₯) and lim π(π₯) exist. π₯→π π₯→π 1. Constant Rule - lim π = π π₯→π 2. Identity Rule - lim π₯ = π π₯→π 3. Sum Rule - lim π π₯ + π(π₯) = lim π(π₯) + lim π(π₯) π₯→π π₯→π π₯→π Limit Laws Let c and k be real numbers so that lim π(π₯) and lim π(π₯) exist. π₯→π π₯→π 4. Difference Rule - lim π π₯ − π(π₯) = lim π(π₯) − lim π(π₯) π₯→π π₯→π 5. Constant Multiple rule - lim π . π π₯ π₯→π π₯→π = π. lim π(π₯) π₯→π 6. Product Rule - lim π π₯ . π(π₯) = lim π(π₯) . lim π(π₯) π₯→π π₯→π π₯→π lim π(π₯) π(π₯) π₯→π 7. Quotient Rule - lim = , where lim π(π₯) ≠ 0 lim π(π₯) π₯→π π(π₯) π₯→π π₯→π 8. Power rule: If n is a positive integer lim π(π₯) π = lim π(π₯) π₯→π π π₯→π 9. Root Rule: If n is a positive integers lim π₯→π π π(π₯) = π lim π(π₯) π₯→π Example Evaluate lim (4π₯2 + 5π₯ − 4) π₯→−2 = lim 4x2 + lim 5x - lim 4 π₯→−2 π₯→−2 π₯→−2 = 4 . lim x2 + 5 . lim x - lim 4 π₯→−2 π₯→−2 = 4 (-2)2 + 5(-2) - 4 = 16 – 10 - 4 =2 π₯→−2 Example π₯ Evaluate lim 2 π₯→3 π₯ +5 = lim π₯ π₯→3 2 lim (π₯ +5) π₯→3 = lim π₯ π₯→3 lim π₯ 2+lim 5 π₯→3 π₯→3 = 3 3 2+5 3 3 = = 9+5 14 Your Turn! Evaluate lim π π₯ + π(π₯) , Given that f(x) = x2 +4x + 4; g(x) = x - 2 π₯→3 lim π π₯ + π(π₯) = x2 +4x + 4 + x - 2 π₯→3 lim π π₯ + π(π₯) = (3)2 +4(3) + 4 + 3 - 2 π₯→3 lim π π₯ + π(π₯) = 9 + 12 + 4 + 3 - 2 π₯→3 lim π π₯ + π(π₯) = 26 π₯→3 Evaluating Limits of functions using other techniques Some limits cannot be evaluated simply using substitution. Particularly in cases when the given is a rational function, direct substitution sometimes yield a number where both the numerator and denominator are zero (0) this may lead into indeterminate form. Indeterminate form are those that cannot be determined by simple evaluating the limits f individual functions. Factoring Method We will factor the given function to avoid indeterminate form. π₯2 − 9 lim π₯→3 π₯ − 3 32 − 9 0 lim = π₯→3 3 − 3 0 This is an indeterminate quantity, which can be circumvented algebraically to evaluate limit through factoring. We will factor the given function to avoid indeterminate form. π₯2 − 9 lim π₯→3 π₯ − 3 (π₯ + 3)(π₯ − 3) lim π₯→3 π₯−3 lim π₯ + 3 π₯→3 lim 3 + 3 π₯→3 6 Rationalization Method 0 Sometimes, the indeterminate form can result from a 0 rational function that contains radical expressions. Cases like this can be circumvented using RATIONALIZATION. π₯−9 lim , ππ¦ π π’ππ π‘ππ‘π’π‘πππ πππ‘βππ π₯→9 π₯ − 3 9−9 lim π₯→9 9 − 3 0 0 The indeterminate form resulted from the radical expression in the denominator. We can circumvent this by rationalization method π₯−9 lim π₯+3 π₯→9 π₯ − 3 π₯−9 . π₯−3 π₯+3 π₯+3 π₯−9 π₯+3 π−9 lim π₯ + 3 π₯→9 lim 9 + 3 π₯→9 =6 SFAC – Taguig Campus Prayer Limits and Infinity Basic Calculus Limits and Infinity Unit I – Module 2 Grade 11 – Basic Calculus SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus One-side Limit The previous discussion revolved on the computation of limit values coming from two directions – from the left and from the right. One condition for the limit of a function as x approaches c from the left is equal to the value of a function as x approaches c from the right. Thus, the expression denotes a two-sided limit. One-sided limits are limits of a function as x approaches c from either the left-hand side or right-hand side. SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Left-hand and Right-hand Limits The limit of function f at x=c as x approaches c from the left side is called the LEFT-HAND LIMIT. Otherwise, it is called RIGHT-HAND LIMIT. SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Left-hand and Right-hand Limits Let the function defined by open interval (x,c) such that c > x. if f approaches the number L as x approaches c within (x,c), then L is called the Left-hand limit of f at c. lim− π π = π³ π→π Let the function defined by open interval (c,x) such that c < x. if f approaches the number M as x approaches c within (c,x), then L is called the Right-hand limit of f at c. lim+ π π = π³ π→π SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example 1 Consider the function π π± = π ππ − π, π ≥ .Evaluate π₯π’π¦+ π(π) π π π→π 1 The expression limit f(x) represents the limit of function as x approaches 2 from the right. Thus, the limit can be calculated by assigning values greater than but arbitrary close to 0.5. x 0.50000001 0.5000001 0.500001 0.50001 0.5001 0.501 f(x) 0.0002 0.0006 0.0020 0.0063 0.0200 0.0632 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calulus x 0.50000001 0.5000001 0.500001 0.50001 0.5001 0.501 f(x) 0.0002 0.0006 0.0020 0.0063 0.0200 0.0632 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example 2 ππ + ππ, π ≤ π Consider the following function: π‘ π = α π π Evaluate lim β(π₯) π + π, π > π π₯→2 Solution: The value of lim β(π₯) will only exist if lim− β(π₯) = lim+ β(π₯) π₯→2 π₯→2 π₯→2 For πππ− π(π) , Substitute the x = 2 in the first piece of the function π→π x2 + 4x (2)2 + 4(2) = 12 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example 2 ππ + ππ, π ≤ π Consider the following function: π‘ π = α π π Evaluate lim β(π₯) π + π, π > π π₯→2 For πππ+ π(π) , Substitute the x = 2 in the second piece of the function π→π 3 π₯+9 2 3 2 +9 2 =12 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Infinite Limits Several example of limit does not exist. Some of these occurs when, after directly substituting the value of c, f(c) is undefined. Usually, in this case, the limit does not exist. However, it is sometime useful to write these limits as infinity (∞) or negative infinity (-∞) SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Consider the function f x 1 = π₯ SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus f x f x 1 = as x approaches 0 from the right π₯ x 0.1 0.01 0.001 0.0001 0.00001 0.000001 y 10 100 1 000 10 000 100 000 1 000 000 1 = as x approaches 0 from the left π₯ x -0.1 -0.01 -0.001 -0.0001 -0.00001 -0.000001 y -10 -100 -1 000 -10 000 -100 000 -1 000 000 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example π Evaluate lim π π→π π By substitution 4 4 πππ 3 = = undefined 0 π₯→0 π₯ You need to reconcile whether it is a positive or a negative infinity. SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example π Evaluate lim π π→π π 4 πππ 3 π₯→0 π₯ 4 = (−0.00001)3 = - 4 x 1015 This means that as x approaches 0 from the left the value of limit is -4 x 1015 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example −πππ Evaluate lim − π π→−π π −π By substitution −πππ lim − π π→−π π − π −π(−π)π = (−π)π −π −6 0 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example −πππ Evaluate lim − π π→−π π −π To get the specified limit, let x be sufficiently close to -1 from the left. x = -1.0001 −πππ lim − π π→−π π − π −π(−π.ππππ)π = (−π.ππππ)π −π Ν -30 004.5 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example −πππ Evaluate lim + π π→−π π −π By substitution −πππ lim + π π→−π π − π −π(−π)π = (−π)π −π −6 0 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example −πππ Evaluate lim + π π→−π π −π To get the specified limit, let x be sufficiently close to -1 from the right. x = -0.9999 −πππ lim − π π→−π π − π −π(−π.ππππ)π = (−π.ππππ)π −π Ν 29 995.5 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Limits at Infinity Infinite limits give information on the existence of vertical asymptotes on the graph of the function. Limits at infinity, on the other hand, are used to show the existence of horizontal. Horizontal Asymptotes are the horizontal lines that a curve approaches as it goes to infinity. SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Consider the function f x 1 = π₯ SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus f x f x 1 = as x approaches 0 from the right π₯ x 0.1 0.01 0.001 0.0001 0.00001 0.000001 y 10 100 1 000 10 000 100 000 1 000 000 1 = as x approaches 0 from the left π₯ x -0.1 -0.01 -0.001 -0.0001 -0.00001 -0.000001 y -10 -100 -1 000 -10 000 -100 000 -1 000 000 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus 1 As x increases without bound positive value, gets sufficiently π₯ 1 close to 0. Thus limits lim = 0. In the same way, as x approaches π₯→∞ π₯ 1 infinity through negative values, also gets sufficiently close to 0. π₯ 1 Therefore, lim = 0. The following statement summarizes this π₯→∞ π₯ observation. SFAC – Taguig Campus Objectives Limits and Infinity Bsic Calculus 1 The Limit of at infinity π₯ 1 As x approaches infinity from either sides, the value of f(x) = π₯ approaches 0. in symbols, you have π πππ = π π→±∞ π SFAC – Taguig Campus Objectives Limits and infinity Basic Calculus Example 3 Evaluate lim π₯→∞ π₯ Solution: Apply the limit laws 3 lim π₯→∞ π₯ 1 = lim 3 . π₯ π₯→∞ 1 = 3 . lim π₯→∞ π = 3 .0 =0 SFAC – Taguig Campus Objectives Limits and Infinity Basic Calculus Example 5 Evaluate lim 2 π₯→−∞ π₯ Solution: Apply the limit laws 5 lim 2 π₯→−∞ π₯ 1 = lim 5 . 2 π₯ π₯→∞ 1 1 = 5 . lim . lim π₯→∞ π π₯→∞ π = 5 .0.0 =0 Objectives SFAC – Taguig Campus Limits and Infinity Basic Calculus 4π₯ 2 −3 Evaluate lim π₯→∞ π₯+4 Example Solution: Apply the limit laws 4π₯ 2 −3 lim π₯→∞ π₯+4 4π₯2 3 − π₯2 π₯2 π₯ 4 π₯→∞ 2 + 2 π₯ π₯ = lim 3 π₯ 1 4 π₯→∞ π₯+ 2 π₯ = lim = 4− 2 3 lim 4− lim 2 π₯→∞ π₯→∞π₯ 1 4 lim π₯+ lim 2 π₯→∞ π₯→∞π₯ 4−0 = 0+0 4 = 0 The limit increases without bound through possible values of x. Thus, 4π₯ 2 −3 lim =∞ π₯→∞ π₯+4 Limits of Exponential, Logarithmic, and Trigonometric Functions UNIT I MODULE 3 GRADE 11 | BASIC CALCULUS Limits of Exponential Function UNIT I MODULE 3 GRADE 11 | BASIC CALCULUS Exponential Function • • • is a function in which the exponent of the expression is a variable that can be written in the form π¦ = π π₯ , π€βπππ b > 0, b ≠ 1. The function is an increasing exponential function when b>1. It is a decreasing exponential function when 0<b<1. Limit of the General Exponential π₯ Function π¦ = π • • • For all real numbers c, π₯π’π¦ π = π π π π→π If b > 1, then π₯π’π¦ π = ∞ and πππ π = π π π→∞ π π→−∞ If 0 < b < 1, then π₯π’π¦ π = π and πππ ππ = ∞ π→−∞ π π→∞ π₯ Example 1. Evaluate lim 2 π₯→3 π₯ lim 2 π₯→3 Example 3. Evaluate lim 2 π₯→∞ lim 2 π₯→∞ π₯ π₯ 1 π₯+2 Example 4. lim π₯→∞ 2 The Natural Exponential Function The exponential function with base e is frequently used in advanced mathematics. π π₯ =π π₯ where: Euler number = 2.718281… Limits of Logarithmic Function LIMITS Logarithmic Function • is the inverse of an exponential function π¦ = π that can be written in the form π¦ = logbπ₯ π₯ For b > 0 and b ≠ 1, the logarithmic function π¦ = logbπ₯ is equivalent to π¦ π₯=π . Limits of Logarithmic Function y = πππππ • For all positive real numbers c in the domain of y = πππππ₯, then lim πππππ₯ = logbc. π₯→π • If b > 1, then lim ππππ π₯ = ∞ and lim+ ππππ π₯ = −∞ • If 0 < b < 1, then lim ππππ π₯ = −∞ and lim+ ππππ π₯ = ∞ π₯→∞ π₯→∞ π₯→0 π₯→0 Example 1. Evaluate lim πππ2π₯ π₯→32 Example 2. Evaluate lim1 πππ32π₯ π₯→6 Limits of Trigonometric Function LIMITS Trigonometric Functions • These include six important functions: sine (sin) cosine (cos) tangent (tan) • • cosecant (csc) secant (sec) cotangent (cot) The sinusoidal trigonometric functions (y=sinx and y=cosx) are considered the most fundamental of all trigonometric functions. Limit of sin x and cos x Because y=sinx and y=cosx are continuous functions, which means they are defined for any real number x, then the limit as x approaches a certain number may be evaluated using direct substitution. - - lim π πππ₯ = π πππ π₯→π lim πππ π₯ = πππ π π₯→π Evaluate the following limits. 1. lim cos π₯ π₯→π 2. lim sin π₯ π₯→π MODULE 4 CONTINUITY A function is said to be continuous on the interval when the function is continuous at a number in the interval. If c is a number in the interval (a, b), and f is a function with a domain containing the interval (a, b), then f is said to be continuous at x = c if all the following conditions are satisfied. What does “continuity at a a) f(c) is defined point” mean? b) lim f x exist x→c c) lim π π₯ = π(π) π₯→π A function is said to be continuous on the interval when the function is continuous at a number in the interval. If c is a number in the interval (a, b), and f is a function with a domain containing the interval (a, b), then f is said to be continuous at x = c if all the Graphically illustrating, a function is following conditions are satisfied. continuous when its graph is a single a) f(c) is defined unbroken curve or line. b) lim f x exist x→c c) lim π π₯ = π(π) π₯→π Example: Discontinuous Continuous A function is said to Function be continuous on the interval whenFunction the function is continuous at a number in the interval. If c is a number in the interval (a, b), and f is a function with a domain containing the interval (a, b), then f is said to be continuous at x = c if all the following conditions are satisfied. a) f(c) is defined b) lim f x exist x→c c) lim π π₯ = π(π) π₯→π CONTINUITY OF A FUNCTION AT A NUMBER A function π π₯ is said to be continuous at π₯ = π, if and only if the following three conditions are satisfied: a. π π is defined; b. πππ π π exist; and π→π c. π π = πππ π π π→π If at least one of these conditions is not met, f is said to be discontinuous at π₯ = π. Example 1: Continuity of Polynomial Function 72 Determine if π π₯ = π₯ 3 + π₯ 2 − 2 is continuous or not at π₯ = 1. Solution: Check the three conditions for the continuity of a function. a. c. find the value of π π determine if π π = πππ π π₯ π₯→π b. determine if πππ π π₯ π₯→π exist Example 2: Continuity of Rational Function Determine if π π₯ 73 π₯ 2 −π₯−2 = is continuous or not at π₯ = 0. π₯−2 Solution: Check the three conditions for the continuity of a function. c. a. find the value of π π b. determine if πππ π π₯ π₯→π exist determine if π π = πππ π π₯ π₯→π CONTINUITY ON AN INTERVAL A function π π₯ can be continuous on an interval. This simply means that it is continuous at every point on the interval. a. Continuity on an Open Interval β’ A function π π₯ is continuous on an open interval (a, b) if it is continuous at every point on the interval (a, b). CONTINUITY ON AN INTERVAL b. Continuity on a Closed Interval β’ A function π π₯ is continuous on a closed interval [a, b] if: 1. 2. It is continuous on the open interval (a, b). It is continuous from the right at a. a. π π ππ₯ππ π‘π . b. πππ+ π π₯ exists. c. 3. π₯→π πππ+ π π₯ = π π . π₯→π It is continuous from the left at b. a. π π ππ₯ππ π‘π . b. πππ− π π₯ exists. c. π₯→π πππ− π π₯ = π π . π₯→π Here are known facts on continuities of functions on intervals: a) Polynomial functions are continuous at all real numbers. b) The absolute value function f(x) = βπ₯βis continuous at all real numbers. c) Rational functions are continuous at any real number that makes the denominator non-zero. d) The square root function f(x) =√π₯ is continuous on (0, ∞). Example 3: Continuity of a Function on an Interval Determine all intervals wherein the function is continuous. Example 4: Continuity of a Function on an Interval Determine whether π π₯ = π₯ 2 + π₯ − 2 is continuous on the interval (−∞, +∞) or not. Solution: Since a polynomial is continuous at every real number, 2 π π₯ = π₯ + π₯ − 2 is continuous at (−∞, +∞). Example 5: Continuity of a Function on an Interval Determine whether π π₯ = 9 − π₯ 2 is continuous on the interval [-3, 3] or not. Solution: DISCONTINUITY Suppose that π is a function defined on the open interval π, π except possibly at the number π contained in π, π . If π is not continuous at π , then π is said to have a discontinuity. TYPES OF DISCONTINUITIES Removable Discontinuity Non-removable Discontinuity TYPES OF DISCONTINUITIES 1. Removable/Hole Discontinuity β’ exists when it can be fixed by redefining the function at the point of discontinuity. β’ represented by a hole on the graph of the function. 2. Non-removable Discontinuity β’ exists when the function cannot be redefined at x=c to make the function continuous. β’ usually from the functions wherein the limit at c does not exist. TWO (2) TYPES OF NON-REMOVABLE DISCONTINUITIES 1. Jump/Essential Discontinuity β’ it happens when the limit does not exist at c because the left-hand limit and right-hand limit are not equal. 2. Infinite/Asymptotic Discontinuity β’ it is represented by a vertical asymptote on the graph and is common to rational functions that cannot be further simplified and written as polynomial functions. Example 6: Discontinuity of a Function Classify the discontinuity of the function π π₯ Solution: 3π₯ 2 +6π₯ = . π₯+2 Example 8: Discontinuity of a Function Classify the discontinuity of the function π π₯ 3 = . 2π₯−1 Solution: 1 The function is not continuous at π₯ = . Because the 2 rational function cannot be further written to its equivalent polynomial function, then the function will have a vertical 1 asymptote at π₯ = . 2 1 Thus, the function has an essential discontinuity at π₯ = . 2
