Differential Calculus - Functions and Limits Lecture Notes PDF

DIFFERENTIAL CALCULUS Topic 4: Functions and Limits, Part 2 Definition of a Limit 𝑓(5.1) = 2(5.1) + 1 𝑓(5.5) = 2(5.5) + 1 Let f be a function defined on some open 𝒇(𝒙) = 𝟏𝟏. 𝟐 𝒇(𝒙) = 𝟏𝟐...

DIFFERENTIAL CALCULUS Topic 4: Functions and Limits, Part 2 Definition of a Limit 𝑓(5.1) = 2(5.1) + 1 𝑓(5.5) = 2(5.5) + 1 Let f be a function defined on some open 𝒇(𝒙) = 𝟏𝟏. 𝟐 𝒇(𝒙) = 𝟏𝟐 interval containing the number a except 𝑓(5.2) = 2(5.2) + 1 𝑓(6) = 2(6) + 1 possibly at a itself. Then the limit of f as x 𝒇(𝒙) = 𝟏𝟏. 𝟒 𝒇(𝒙) = 𝟏𝟑 approaches a is L, written as: 𝑓(5.3) = 2(5.3) + 1 𝑓(6.5) = 2(6.5) + 1 𝒇(𝒙) = 𝟏𝟏. 𝟔 𝒇(𝒙) = 𝟏𝟒 lim 𝑓(𝑥 ) = 𝐿 𝑓(5.4) = 2(5.4) + 1 𝑓(7) = 2(7) + 1 𝑥→𝑎 if the value of f gets closer and closer to one 𝒇(𝒙) = 𝟏𝟏. 𝟖 𝒇(𝒙) = 𝟏𝟓 and only one number L as x takes values that are closer and closer to a. Right-side x 5+ 5.1 5.2 5.3 5.4 5.5 6 6.5 7 NOTE: The Limit of a function refers to the y value that the function approaches, not the f(x) 11.2 11.4 11.6 11.8 12 13 14 15 actual value (if any) lim 𝑓(𝑥 ) = 𝐿 𝑥→𝑐 read as “limit of f(x) as x approaches c is equal to L” Example: lim (2𝑥 + 1) Note: if the limit of the function from left to 𝑥→5 right are different, then the limit does not exist 𝑓(𝑥) = 2𝑥 + 1 𝑓(4.7) = 2(4.7) + 1 𝑓(3) = 2(3) + 1 𝒇(𝒙) = 𝟏𝟎. 𝟒 𝒇(𝒙) = 𝟕 𝑓(4.8) = 2(4.8) + 1 𝑓(3.5) = 2(3.5) + 1 𝒇(𝒙) = 𝟏𝟎. 𝟔 𝒇(𝒙) = 𝟖 𝑓(4.9) = 2(4.9) + 1 𝑓 (4 ) = 2 (4 ) + 1 𝒇(𝒙) = 𝟏𝟎. 𝟖 𝒇(𝒙) = 𝟗 𝑓(4.5) = 2(4.5) + 1 𝒇(𝒙) = 𝟏𝟎 𝑓(4.6) = 2(4.6) + 1 𝒇(𝒙) = 𝟏𝟎. 𝟐 Left-side x 5− 3 3.5 4 4.5 4.6 4.7 4.8 4.9 5 y f(x) 7 8 9 10 10.2 10.4 10.6 10.8 Theorem and Proof on Limit Theorem 5: Product Rule Suppose 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳, and 𝐥𝐢𝐦 𝒈(𝒙) = 𝑴, 𝒙→𝒂 𝒙→𝒂 Theorem 1: Where c is constant then; 𝐥𝐢𝐦[𝒇(𝒙) ∙ 𝒈(𝒙)] = 𝑳 ∙ 𝑴 𝐥𝐢𝐦 𝒄 = 𝒄 𝒙→𝒂 𝒙→𝒂 Ex: Ex: 1. lim 5 2. lim(−2) lim 𝑥 (6𝑥 − 2) 𝑥→4 𝑥→0 𝑥→3 =5 = -2 = lim 𝑥 ∙ lim (6𝑥 − 2) 𝑥→3 𝑥→3 = lim 𝑥 ∙ [lim 6𝑥 − lim 2] 𝑥→3 𝑥→3 𝑥→3 Theorem 2: for any number a: = 3 ∙ [lim 6𝑥 − lim 2] 𝐥𝐢𝐦 𝒙 = 𝒂 𝑥→3 𝑥→3 𝒙→𝒂 = 3[6(3) − 2] Ex: = 3 1. lim 𝑥 2. lim1 𝑥 𝑥→4 𝑥→3 = 𝟒𝟖 𝟏 =4 = 𝟑 Theorem 6: Quotient Rule Suppose 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳, and 𝐥𝐢𝐦 𝒈(𝒙) = 𝑴, 𝒙→𝒂 𝒙→𝒂 Theorem 3: Constant Multiple Rule 𝒇(𝒙) 𝑳 then; 𝐥𝐢𝐦 𝒈(𝒙) = 𝑴 Suppose 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳, then; 𝒙→𝒂 𝒙→𝒂 Ex: 𝐥𝐢𝐦[𝒌 ∙ 𝒇(𝒙)] = 𝒌 ∙ 𝑳 where k is constant 𝑥 𝒙→𝒂 lim 𝑥 𝑥→−2 𝑥−4 Ex: lim 𝑥 𝑥→−2 lim 5𝑥 = 𝑥→3 lim (𝑥 − 4) 𝑥→−2 = 5lim 𝑥 −2 𝑥→3 = lim 𝑥 − lim 4 = 5(3) 𝑥→−2 𝑥→−2 −2 = 𝟏𝟓 = −2 − 4 −2 = Theorem 4: Sum/Difference Rule −6 𝟏 Suppose 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳, and 𝐥𝐢𝐦 𝒈(𝒙) = 𝑴 = 𝒙→𝒂 𝒙→𝒂 𝟑 then; 𝐥𝐢𝐦[𝒇(𝒙) ± 𝒈(𝒙) = 𝑳 + 𝑴 𝒙→𝒂 Theorem 7: Power Rule Ex: 𝐥𝐢𝐦[𝒇(𝒙)]𝒑 = [𝐥𝐢𝐦 𝒇(𝒙)]𝒑 𝒙→𝒂 𝒙→𝒂 lim 2𝑥 = 4 𝑥→1 = lim 2𝑥 + lim 4 𝑥 Ex: 𝑥→1 𝑥→1 lim 𝑥 3 = 2lim 𝑥 + 4 lim 𝑥 𝑥→2 𝑥→1 𝑥→1 = [lim 𝑥]3 = 2(1) + 4(1) 𝑥→2 3 =𝟔 = =𝟖 Theorem 8: Radical/Root Rule Right-Hand Limit We say 𝐥𝐢𝐦+ 𝒇(𝒙) = 𝑳 provided; we can 𝐥𝐢𝐦 √𝒙 = √𝐥𝐢𝐦 𝒙 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂 make f(x) as close to L as we want for x Ex: sufficiently close to a with x > a without lim √𝑥 actually letting x be a. 𝑥→9 = √9 Left-Hand Limit =𝟑 We say 𝐥𝐢𝐦+ 𝒇(𝒙) = 𝑳 provided we can 𝒙→𝒂 make f(x) as close to L as we want for x Theorem 9: Limit of a Polynomial sufficiently close to a with x < a without Function actually letting x be a. If p(x) and q(x) are polynomials, then; 𝐥𝐢𝐦 𝒑(𝒙) = 𝒑(𝒂) 𝒙→𝒂 𝒑(𝒙) 𝒑(𝒂) 𝐥𝐢𝐦 𝒒(𝒙) = 𝒒(𝒂) if q(a)≠0 𝒙→𝒂 Ex: lim(𝑥 2 − 3𝑥 + 5) 𝑥→3 = 𝑓(3)2 − (3)(3) + 5 =9−9+5 =5 Right-Hand and Left-Hand Limit Once in a while it is convenient to employ a restricted version of limit as described below. We write; 𝑓(𝑥)=L And mean by 𝒙→𝒂^+that each x involved is greater than a. A limit such as that in (1) is called a right-hand limit; the independent variable x approaches a from the right. A left-hand limit (2) 𝑓(𝑥)=𝑀 with x remaining less than a, is also used. If the ordinary limit exists, the right-hand and left-hand limits each exist and all three have the same value. If the right- and left-hand limits exist and have the same value, the limit itself exists and has that value.