Could someone please check my work

ase state all definitions and theorems that you will need: Definition 5.1.12 Let L = lim f(x) = lim 1 -4= - 3 Let f: D - R and g: D -+ R . We define the sum f + g and the product fg to be the functions and from D to R given by M = lim g(x) = lim 1 -5 = -4 (f + 9)(z) = f(z) + g(z) and (fg)(z) = f(x) - 9(z) for all I E D. If k E R, then the multiple kf : D -> R is the function defined by Then by the quotient formula of theorem 5.1.13 and definition 5.1.12 for the quotient of -:D - R, (kf)(x) = k . f(x) for all z E D. since M # 0. If g(z) * 0 for all at E D , then the quotient - : D -+ R is the function defined by (4 ) (2) = 121, for all z ED. Theorem 5.1.13 Let f : D - R and g: D + R, and let c be an accumulation point of D . If lim f(x) = L, VI - 1 3. limo -1 = 2 lim g(x) = M, and & E R , then Explain: lim (f + 9)(x) = L+ M, lim (fg)(x) = LM, Since- Va - 1 Va - 1 and lim (kf) (x) = KL . I - I ( Vz - 1) (Vi+1) Va+1 Furthermore, if g() * 0 for all x E D and M # 0, then Let f(x) = 1 and g(x) = Vx + 1 and c = 1 where f: D -+ R and 9: D - R. lim (2) = M Since g(z) = Vi + 1 / 0 for all x E D because va + 1 is always positive, then by definition 5.1.12 for the quotient of ( ) (x) : D - R, (! ) ( 20 ) = 1(2 ) Vit I Let L = lim f(x) = lim 1 =1 and Determine the following limits. M = lim g(x) = lim VI+1 =2 25 + 4 1. lim 2 + 1 Then by the quotient formula of theorem 5.1.13 and definition 5.1. 12 for the quotient of - : D - R, Explain: since g(2) # 0 for all x E D and M # 0, Let f(2) = 2 + 4 and g(x) = 22 + 1 and c = 1 where f : D - R and g: D - R. lim ((2) = M = 2 Since g(z) = x2 + 1 / 0 for all a E D because x2 + 1 is always positive, then by definition 5.1.12 for the quotient of ( ) (z) : D - IR, =f(2) (" ( 2 ) = 2 25 + 4 g (z ) *2 + 1 Let L = lim f(x) = lim 13 + 4 = 5 and M = lim g(x) = lim 12 + 1 = 2 Then by the quotient formula of theorem 5.1.13 and definition 5.1.12 for the quotient of -: D - R, since g(x) # 0 for all x E D and M # 0, lim ((2) = M = 2 2. lim - 24 - 5x + 4 241 x2 - 62 + 5 Explain: nce 27 - 5x + 4 (2 - 1)(2 - 4) 2 - 4 x2 - 6x + 5 (x - 1)(x -5) I - 5 Let f(x) = x - 4 andg(x) = x - 5 and c = 1 where f : D - IR and g: D - R. Since g (x) = x - 5 = 0 when a = 5 but we are only interested in the limit at a = 1, then by definition 5.1.2 for the quotient of () (2) : D - R