Could someone please check my work

Tht'on'm 5 1 8 Let f; I) ~ '4 and let 1' be an accumulation point of I) . Then lim f(.r) I. Hi for every I 'l' sequence (x, 'l in I) that converges to r with 3,, 7 v for all n. the sequence (f(.4,,)) converges to 1.. Theorem 4.1.l4 H a sequenu' converges. its limit IS unique. Prove Corollary 5.1.9: if f: I? x R and if c is an accumulation point of D , then f can have only one limit at c . 'r' u. ,. ' ' 2. by using Theorems 5.1.8 and 4.1.14. Suppose f: D + R and c is an accumulation point of D . Assume lim f(a:) = L . :L'->c Then by Theorem 5.1.8, there exists a sequence, (3\") E D such that (37,) > 050 that (f(sn)) ~> L where 3,; 7E cfor all 'n E N , which means lim f(.sn) = Lfor all 3,, E (8\") E D. n>oo Now suppose f has a second limit point, M. Then by the same theorem 5.1.8, Then by theorem 5.1.8, since a convergent sequence can have only one limit, L = M . Hence, it is not the case that there are tw0 limits, which means we've reached a contradiction. Therefore, if f: D ~+ R and c is an accumulation point of D , then f can have only one limit at c . In other words, the limit of a sequence of functions, if it exists, is unique. ____