Could someone please check my work I have to use the included theorems

finitions and theorems that you will need: Theorem 4.2.1 Suppose that (s.) and (. ) are convergent sequences with Tim s, "'s and lim , " . Then (by tim (2 .3. ) - ks and tim (2 + 3.) - 4 + s. for any & E R (6) tim - ; . provided that , # 0 for all * and : 2 0 Theorem 5.1.8 Let : D) - R and let c be an accumulation point of D . Then lim f(x) - Z Iff for every sequence (s. ) in D that converges to e with s, # e for all n, the sequence ((s, ) ) converges to L. Definition 5.1.12 Let f: D - R and g: D - R . We define the sum f+ g and the product fg to be the functions from D to R given by (+ 8)(x) = (x) + 8(x) and (2)(x) - Ax) . g(x) for all x E D . If & E R , then the multiple Af: D -. R is the function defined by (A/)(x) - A . Rx) for all x ED. If g(x) # 0 for all x E D , then the quotient - : D - R is the function defined by (() - , for all. ED . -..... Let f: D - R and f: D - R , and let c be an accumulation point of D . x -C If lim f(x) = L, lim g(x) = M , and k E R , then X - C 1. lim (g) (x) = LM , Let (Sm) be a sequence in D that converges to c with each s,, # c . By Theorem 5.1.8, we have 1 - 00 lim ($, ) = L and lim 8(57) = M, 1 - + 00 1 - 00 and it suffices to show that lim (f . g) (S, ) = L . M . Now from definition 5.1.12 and theorem 4.2.1 we obtain 1-+ 00 lim (fg) (5, ) = lim [(s,) 8(s, ) ] = lim f(Sn ) . lim g(5 ) 1 - 00 = L . M 2. lim (k)(x) = kL . x -C Let (57 ) be a sequence in D that converges to c with each s,, # c . By Theorem 5.1.8, we have lim 1 ( S , ) = L , 1- 00 and it suffices to show that lim (k . D)(S, ) = k . L . Now from definition 5.1.12 and theorem 1-+ 00 4.2.1 we obtain lim (k . D)(s,, ) = k . lim f(s;) 1 -+ 90 m k . L 3. Furthermore, if g(x) # 0 for all x E D and M * 0, then lim ( )() - k. Suppose g (57 ) * 0 for all s,, E D and all n E N and suppose M # 0 . Let (s) be a sequence in D that converges to c with each s,, A c . By Theorem 5.1.8, we have lim (S. ) = L and lim &(5, ) = M, and it suffices to show that lim ()(s.) - M. Now from definition 5.1.12 and theorem 4.2.1 we obtain lim ()() - lima [(") "- x (s ). lim /( SM) lim g ($,)