Galala University Mathematics I (MAT 111) Lecture 4 PDF

Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 4 The Limit of a Function_Part 1 Dr. Sameh Basha ...

Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 4 The Limit of a Function_Part 1 Dr. Sameh Basha  Introduction to limits 𝑥−1 Let 𝑓 𝑥 =. 𝑥 2 −1 Suppose we need to find 𝑓 1 𝑥−1 Notice that: 𝑓 𝑥 = not defined when x=1. 𝑥 2 −1 So, let us investigate the behavior of the function for values of x near to 1. near to 1 𝒙𝟏 Dr. Sameh Basha  near to 1 𝒙𝟏 lim 𝟐 = 𝟎. 𝟓 𝒙→𝟏 𝒙 − 𝟏 Dr. Sameh Basha  near to 1 𝒙𝟏 lim 𝟐 = 𝟎. 𝟓 𝒙→𝟏 𝒙 − 𝟏 The closer 𝒙 is to 𝟏 The closer 𝐟(𝒙) is to 𝟎. 𝟓 𝒙−𝟏 → 𝟎. 𝟓 As 𝒙 → 𝟏 𝒙𝟐 −𝟏 Dr. Sameh Basha  Now, Let us investigate the behavior of the function 𝑓 𝑥 = 𝑥 2 − 𝑥 + 2 for values of 𝑥 near to 2. Near to 2 𝒙𝟐 𝒙→𝟐 Dr. Sameh Basha Intuitive Definition of a Limit Dr. Sameh Basha Note Let: 𝑥 2 ; 𝑖𝑓 𝑥 > 0 f 𝑥 =ቊ 5; 𝑖𝑓 𝑥 = 0 𝐷𝑓 = 0, ∞ 𝑓 0 =? ? 𝑓 0 =5 Dr. Sameh Basha  Note: 𝒇(𝒙) is defined 𝒇(𝒙) is defined 𝒇(𝒙) is Not defined at 𝐱 = 𝒂. at 𝐱 = 𝒂. at and and 𝐱 = 𝒂. 𝒇 𝒂 =𝑳 𝒇 𝒂 =𝒃 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 = 𝒇(𝒂) 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 𝒙→𝒂 𝒙→𝒂 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 𝒇 𝒂 =𝒃 𝒙→𝒂 Dr. Sameh Basha Note: Dr. Sameh Basha Note: Dr. Sameh Basha  𝒕𝟐 +𝟗−𝟑 Ex: Estimate the value of 𝒍𝒊𝒎 𝒕→𝟎 𝒕𝟐 As t approaches 0, the values of the function seem to approach 0.1666666... and so we guess that: 𝒕𝟐 + 𝟗 − 𝟑 𝟏 𝒍𝒊𝒎 𝟐 = 𝒕→𝟎 𝒕 𝟔 Dr. Sameh Basha What would have happened if we had taken even smaller values of t? Something strange seems to be happening. If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make t sufficiently small. 𝟏 Does this mean that the answer is really 0 instead of ? 𝟔 𝟏 No, the value of the limit is 𝟔 The problem is that the calculator gave false values because 𝒕𝟐 + 𝟗 − 𝟑 is very close to 3 when t is small. In fact, when t is sufficiently small, a calculator’s value for 𝒕𝟐 + 𝟗 − 𝟑 is 3.000... to as many digits as the calculator is capable of carrying. rounding errors from the subtraction. Dr. Sameh Basha  One-Sided Limits 𝟎; 𝒊𝒇 𝒕 < 𝟎 𝑯 𝒕 =ቊ 𝟏; 𝒊𝒇 𝒕 ≥ 𝟎 As t approaches 0 from the left, 𝑯 𝒕 approaches 0. As t approaches 0 from the right, 𝑯 𝒕 approaches 1. There is no single number that 𝑯 𝒕 approaches as t approaches 0. Therefore, 𝒍𝒊𝒎 𝑯 𝒕 does not exist. 𝒕→𝟎 As t approaches 0 from the left, 𝑯 𝒕 approaches 0. As t approaches 0 from the right, 𝑯 𝒕 approaches 1. 𝒍𝒊𝒎− 𝑯 𝒕 = 𝟎 𝒍𝒊𝒎+ 𝑯 𝒕 = 𝟏 𝒕→𝟎 𝒕→𝟎 Therefore, 𝒍𝒊𝒎 𝑯 𝒕 does not exist. 𝒕→𝟎 Dr. Sameh Basha One-Sided Limits 𝒕 → 𝟎− indicates that we consider only values of t that are less than 0. 𝒕 → 𝟎+ indicates that we consider only values of t that are greater than 0. Definition: Definition: We write 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑳 We write 𝒍𝒊𝒎+ 𝒇 𝒙 = 𝑳 𝒙→𝟎 𝒙→𝒂 and say the left-hand limit of 𝒇 𝒙 as x approaches a and say the right-hand limit of 𝒇 𝒙 as x approaches a [or the limit of 𝒇 𝒙 as x approaches a from the left] [or the limit of 𝒇 𝒙 as x approaches a from the right] is equal to 𝑳. if we can make the values of 𝒇 𝒙 is equal to 𝑳. if we can make the values of 𝒇 𝒙 arbitrarily close to 𝑳 by taking x to be sufficiently arbitrarily close to 𝑳 by taking x to be sufficiently close to a with x less than a. close to a with x greater than a. Dr. Sameh Basha Definition: Dr. Sameh Basha Ex: The graph of a function t is shown in Figure. Use it to state the values (if they exist) of the following: a) 𝒍𝒊𝒎− 𝒈 𝒙 b) 𝒍𝒊𝒎+ 𝒈 𝒙 c) 𝒍𝒊𝒎 𝒈 𝒙 g) g(5) 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒍𝒊𝒎− 𝒈 𝒙 e) 𝒍𝒊𝒎+ 𝒈 𝒙 f) 𝒍𝒊𝒎 𝒈 𝒙 𝒙→𝟓 𝒙→𝟓 𝒙→𝟓 Answer: a) 𝒍𝒊𝒎− 𝒈 𝒙 = 3 b) 𝒍𝒊𝒎+ 𝒈 𝒙 = 1 c) 𝒍𝒊𝒎 𝒈 𝒙 = Does not Exist 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 g) g(5) = 1 d) 𝒍𝒊𝒎− 𝒈 𝒙 = 2 e) 𝒍𝒊𝒎+ 𝒈 𝒙 = 2 f) 𝒍𝒊𝒎 𝒈 𝒙 = 2 𝒙→𝟓 𝒙→𝟓 𝒙→𝟓 Dr. Sameh Basha Ex: Use the given graph of f to state the value of each quantity, if it exists. If it does not exist. a) 𝒍𝒊𝒎− 𝒇 𝒙 b) 𝒍𝒊𝒎+ 𝒇 𝒙 c) 𝒍𝒊𝒎 𝒇 𝒙 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒇(𝟐) e)𝒍𝒊𝒎 𝒇 𝒙 f)𝒇(𝟒) 𝒙→𝟒 Answer: a) 𝒍𝒊𝒎− 𝒇 𝒙 = 3 b) 𝒍𝒊𝒎+ 𝒇 𝒙 = 1 c) 𝒍𝒊𝒎 𝒇 𝒙 = Does not Exist 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒇 𝟐 = 3 e)𝒍𝒊𝒎 𝒇 𝒙 = 4 f)𝒇 𝟒 = Does not Exist 𝒙→𝟒 Dr. Sameh Basha Infinite Limits 𝟏 Ex: Find 𝒍𝒊𝒎 𝟐 (if it exists) 𝒙→𝟎 𝒙 Answer: As 𝑥 becomes close to 0, 𝑥 2 also becomes close to 0, 1 and becomes very large. 𝑥2 𝟏 lim 𝟐 = ∞ 𝒙→𝟎 𝒙 Dr. Sameh Basha  lim 𝒇(𝒙) = ∞ 𝒙→𝒂 to indicate that the values of 𝒇 𝒙 tend to become larger and larger (or “increase without bound”) as 𝑥 becomes closer and closer to 𝑎. (𝑓(𝑥) → ∞ as 𝑥 → 𝑎 ) Dr. Sameh Basha Infinite Limits 𝟏 Ex: Find 𝒍𝒊𝒎 − 𝟐 (if it exists) 𝒙→𝟎 𝒙 Answer: As 𝑥 becomes close to 0, 𝑥 2 also becomes close to 0, 1 and - 2 becomes very large negative. 𝑥 𝟏 lim − 𝟐 = −∞ 𝒙→𝟎 𝒙 Dr. Sameh Basha  lim 𝒇 𝒙 = −∞ 𝒙→𝒂 to indicate that the values of 𝒇 𝒙 tend to become larger and larger negative (or “decrease without bound”) as 𝑥 becomes closer and closer to 𝑎. (𝑓 𝑥 → −∞ as 𝑥 → 𝑎 ) Dr. Sameh Basha Similar definitions can be given for the one-sided infinite limits: 𝒍𝒊𝒎 𝒇(𝒙) = ∞ 𝒍𝒊𝒎 𝒇(𝒙) = ∞ 𝒙→𝒂− 𝒙→𝒂+ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒙→𝒂− 𝒙→𝒂+ Note that: 𝒙 → 𝒂− means that we consider only values of x that are less than a, 𝒙 → 𝟐+ means that we consider only values of x that are greater than a Dr. Sameh Basha Vertical asymptote Dr. Sameh Basha The line 𝒙 = 𝒂 is a vertical asymptote line in each of the following: Dr. Sameh Basha For the function A whose graph is shown, state the following. 𝒂) 𝒍𝒊𝒎 𝑨 𝒙 𝒃) 𝒍𝒊𝒎+ 𝑨 𝒙 𝒙→−𝟑 𝒙→𝟐 𝒄) 𝒍𝒊𝒎 𝑨 𝒙 𝒅) 𝒍𝒊𝒎 𝑨 𝒙 𝒙→𝟐− 𝒙→−𝟏 e) The equations of the vertical asymptotes. Answer: 𝒂) 𝒍𝒊𝒎 𝑨 𝒙 = ∞ 𝒃) 𝒍𝒊𝒎+ 𝑨 𝒙 = ∞ 𝒙→−𝟑 𝒙→𝟐 𝒄) 𝒍𝒊𝒎 𝑨 𝒙 = −∞ 𝒅) 𝒍𝒊𝒎 𝑨 𝒙 = −∞ 𝒙→𝟐 − 𝒙→−𝟏 e) The equations of the vertical asymptotes. 𝑥 = −3, 𝑥 = −1, 𝑥 = 2 Dr. Sameh Basha Calculating Limits Using the Limit Laws Dr. Sameh Basha Calculating Limits Using the Limit Laws Dr. Sameh Basha Direct Substitution Property Dr. Sameh Basha Ex: Evaluate the following limits (if exists): 𝟐 𝒙𝟐 −𝟑 1)𝒍𝒊𝒎(𝟐𝒙 − 𝟑𝒙 + 𝟒) 2)𝒍𝒊𝒎 𝟐 𝒙→𝟓 𝒙→𝟐 𝒙 +𝟐 Answer: Answer: 𝒍𝒊𝒎 𝟐𝒙𝟐 − 𝟑𝒙 + 𝟒 𝒙𝟐 −𝟑 𝟐𝟐 −𝟑 𝟒−𝟑 𝟏 𝒙→𝟓 𝒍𝒊𝒎 𝟐 = 𝟐 = = = 𝟐 𝟓 𝟐−𝟑 𝟓 +𝟒 𝒙→𝟐 𝒙 +𝟐 𝟐 +𝟐 𝟒+𝟐 𝟔 = 𝟓𝟎 − 𝟏𝟓 + 𝟒 = 𝟑𝟗 𝒙𝟐 −𝟗 𝒙𝟐 −𝟒 3)𝒍𝒊𝒎 𝟐 4)𝒍𝒊𝒎 𝒙→𝟑 𝒙 +𝟑 𝒙→𝟐 𝒙−𝟐 Answer: Answer: 𝒙𝟐 − 𝟗 𝟑𝟐 − 𝟗 𝟗 − 𝟗 𝒛𝒆𝒓𝒐 𝒙𝟐 −𝟒 𝒙−𝟐 𝒙+𝟐 𝒍𝒊𝒎 𝟐 = 𝟐 = = = 𝒁𝒆𝒓𝒐 𝒍𝒊𝒎 𝟐 =𝒍𝒊𝒎 = 𝒍𝒊𝒎 𝒙 + 𝟐 = 𝟒 𝒙→𝟑 𝒙 + 𝟑 𝟑 +𝟑 𝟗+𝟑 𝟏𝟐 𝒙→𝟐 𝒙 −𝟐 𝒙→𝟐 𝒙−𝟐 𝒙→𝟐 Dr. Sameh Basha Note: Dr. Sameh Basha Theorem 1: Dr. Sameh Basha Theorem 2: Dr. Sameh Basha Theorem 3: Squeeze Theorem (Sandwich Theorem) Dr. Sameh Basha Example: Let 𝑓 𝑥 = ቊ 𝑥 − 4; 𝑥 > 4 8 − 2𝑥; 𝑥 < 4 Evaluate 𝒍𝒊𝒎𝒇(𝒙) (if exists) 𝒙→𝟒 Answer: 𝒍𝒊𝒎 𝒇 𝒙 = 𝒍𝒊𝒎+ 𝑥 − 4 = 4 − 4 = 𝑍𝑒𝑟𝑜 = 𝑍𝑒𝑟𝑜 𝒙→𝟒+ 𝒙→𝟒 𝒍𝒊𝒎− 𝒇 𝒙 = 𝒍𝒊𝒎− 8 − 2𝑥 = 8 − 2 4 = 8 − 8 = 𝑍𝑒𝑟𝑜 𝒙→𝟒 𝒙→𝟒 ∵ 𝒍𝒊𝒎 + 𝒇 𝒙 = 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑍𝑒𝑟𝑜 𝒙→𝟒 𝒙→𝟒 ∴ 𝒍𝒊𝒎 𝒇(𝒙) = 𝒁𝒆𝒓𝒐 𝒙→𝟒 Dr. Sameh Basha Example: Evaluate (if exists) 𝒍𝒊𝒎𝒇(𝒙) where, 𝒙→𝟎 𝑓 𝑥 = 𝑥 Answer: 𝒙; 𝒙≥𝟎 𝒇 𝒙 = 𝒙 =ቊ −𝒙; 𝒙