Could someone please check my work

Please state all definitions and theorems that you Will need: Definition 5.2.1 L9t f3 D E and let 4* ., D. We say that [is continuous at r' if A. l ' 6 nd i _ . r ,1: Ci '1 a for every 5 \" ll there exists a it \\~ 0 such that Ffl-l'i ' flal '* '3 wheneve I .r ._ D. Theorem 5.7.2 Let f: I) x 3:1 and let r' i, D . Then the following three conditions are equivalent: in) f is continuous at i' lb} If (.r,,) is any sequence in D such that (.r\") converges to c . then Illixtloo it") ' c) (c) For every neighborhood V of f(r') there exists a neighborhood U of c such that fll' n 1)) \\' Furthermore, if c is an accumulation point of D , then the above are all equivalent to (d) f has a limit at c and lim ns) 2 f(c) 1-?(3 Let f: D -+ R be continuous at c E D and suppose that f(c) > 0 . Prove that there exists an a > 0 and a neighborhood U of c such that f(a:) > a for all x E U 0 D . f (2 Let f:D > R be continuous ate 6 D and supposef(c) > 0. Let s = :2) > 0. Since f is continuous at c , then by definition 5.2.1, f (C) _ V>0 36> OsuchthatiED,1mcl |f(a:)f(c)| f(:c) E V(f(c); \"20) by the definition of neighborhood of f (C) - 0 Since f(c) > 0 and 11) > 0 then f(m) > 0. 2 Thus, there exists an a > 0 such that f(:c) > or > 0 . " I Since [:12 c| ac E U(c; 6) by the definition of neighborhood of c, and since Vac E D, |m cl 0 y then by Theorem 5.2.2(c) and definition 5.2.1, there exists an 01 > 0 and a neighborhood U of c such that at) > a for all an E U D D . Therefore, when f: D > R is continuous at c E D and f(c) > 0 , there exists an a > 0 and a neighborhood U of c such that f(a:) > a for all a: E U (1 D