University:
Kentucky State University
Course:
MAT 120 | Precalculus
Academic year:

2021
Views:

220

Pages:

7
Author:

Dixie Summers

Limits 14 Nov 2021 Definition: lim f(x) = l "exists" x→a if lim f(x) = lim f(x) = l x→a⁺ x→a⁻ if lim f(x) ≠ lim f(x) then lim f(x) doesn't exist "dne" x→a⁺ x→a⁻ x→a Ex: Is lim f(x) exist? x→2 lim f(x) = 3 x→2⁺ lim f(x) = 1 ≠ 3 → dne x→2⁻ f(2) = ?? ⇒ f(2) = 2 Theorem: If lim f(x) = M, lim g(x) = L then: x→a x→a lim [f(x) ± g(x)] = lim f(x) ± lim g(x) = M ± L x→a x→a x→a lim [f(x) * g(x)] = lim f(x) * lim g(x) = M * L x→a x→a x→a lim [f(x) / g(x)] = lim f(x) / lim g(x) = M/L, L ≠ 0 x→a x→a x→a lim [f(x)]ⁿ = [lim f(x)]ⁿ = Lⁿ x→a x→a lim C = C, C is constant x→a Ex: If lim f(x) = 3, lim g(x) = 2. Find: x→a x→a lim [f(x) + g(x)] = lim f(x) + lim g(x) x→a x→a x→a = 3 + 2 = 5 2 lim f(x) = 9 = 2/2 = 5 x→2 **Ex: Find** lim (2x² - 4x + 5) = 2(0)² - 4(0) + 5 = 5 x→0 **Note that:** lim f(x) = Number x→a lim g(x) = Number **Ex:** lim (3x+5) = 12+5 = 5 x→4 x+4 √x+4 lim (2x-1) = 1-1 = 0 x→1 2x lim f(x) = 0 x→a lim g(x) = 0 **Ex:** lim x-9 = 0 (x-3)(x+3) = x+3 = 3+3 = 6 x→3 x-3 lim 2x+8 = 2(3+4) = 2 x→3 x+4 lim x²-x-12 = (x-4)(x+3) = 2 x→4 x²-4 lim 5x-5 = 5(2-1) = 5 x→2 x-1 lim x²-x-12 = 5(3+4) - 16 = 5x10-16 = 34 x→2 3(x+4) (5) lim (x²-2)² = x²-4x+4 = 4 x→0 x (6) lim |x-4| = 4-4 = 0 x→4- x-4 lim |x-4| = -(x-4) = -4+4 = 0 x→4⁺ x-4 lim |x-3| = -1 x→3- 2(x-3) = 2 lim √(x+15) - 2 = (√(x+15) - 2)(√(x+15) + 2) = x+15-4 x→1 x-1 (x-1)(√(x+15) + 2) y = √(x+15) - 2 y² = x+15-4 y² = x+11 **Note that:** 1. a²-b² = (a-b)(a+b) 2. a³-b³ = (a-b)(a²+ab+b²) 3. a³+b³ = (a+b)(a²-ab+b²) Ex: lim 2-√(x+3) = 2+√(x+3) = 9-(x+3) = 9-x-3 = 6 x→1 x²+2x-3 x²+2x-3 x²+2x-3 x²+2x-3 4-(x+3) = 4-x-3 = 1-x = 1-1 = 0 (x+1)(x-1) (x+1)(x-1) Ex: lim 3√x+21 - 3 = 3³ + (3²)(√x+21) + (√x+21)² = 27 + 9√(x+21) + x+21 x→9 3√x+9 - 3 3√x+9 + 3 = 27-(2x+21) = 6-2x = 6-2(9) = -12 (x-9)(27+9√(x+21)+x+21) 27(9+9) = -2/27(18) = -1/81 Ex: lim √x-2 = √6-2 = √4 = 2 x→4 x-2 Ex: lim √x-3 = √1-3 = √-2 = d.n.e x→1 x-3 Ex: lim √2x = √2+2 = 0 + 2 = 2 x→2 x-2 lim √x+2x = d.n.e x→-2 lim 1/x = ∞ x→0 Eg) f(x) = {√(x+4), x ≤ 4 {8-2x, x > 4 a) Find lim f(x) = lim 8-2x = 8-2*4 = 0 x→4⁺ x→4⁺ b) Find lim f(x) = lim √(x+4) = √(4+4) = 2 x→4⁻ x→4⁻ c) lim f(x) = 0 ≠ 2 x→4 Eg) f(x) = {x², x ≤ 2 {2x+K, x > 2 Find K such that lim f(x) exists x→2 lim 2x+K = lim x² x→2⁺ x→2⁻ 2*2+K = 2² 4+K = 4 K = 0