Introduction to Mathematical Analysis I - Second Edition

75 Example 3.3.1 Let f : R → R be given by f ( x ) = 3 x + 7. Let x 0 ∈ R and let ε > 0. Choose δ = ε / 3. Then if | x − x 0 | < δ , we have | f ( x ) − f ( x 0 ) | = | 3 x + 7 − ( 3 x 0 + 7 ) | = | 3 ( x − x 0 ) | = 3 | x − x 0 | < 3 δ = ε . This shows that f is continuous at x 0 . Remark 3.3.1 Note that the above definition of continuity does not mention limits. This allows to include in the definition, points x 0 ∈ D which are not limit points of D . If x 0 is an isolated point of D , then there is δ > 0 such that B ( x 0 ; δ ) ∩ D = { x 0 } . It follows that for x ∈ B ( x 0 ; δ ) ∩ D , | f ( x ) − f ( x 0 ) | = 0 < ε for any epsilon. Therefore, every function is continuous at an isolated point of its domain. To study continuity at limit points of D , we have the following theorem which follows directly from the definitions of continuity and limit. Theorem 3.3.2 Let f : D → R and let x 0 ∈ D be a limit point of D . Then f is continuous at x 0 if and only if lim x → x 0 f ( x ) = f ( x 0 ) . Example 3.3.2 Let f : R → R be given by f ( x ) = 3 x 2 − 2 x + 1. Fix x 0 ∈ R . Since, from the results of the previous theorem, we have lim x → x 0 f ( x ) = lim x → x 0 ( 3 x 2 − 2 x + 1 ) = 3 x 2 0 − 2 x 0 + 1 = f ( x 0 ) , it follows that f is continuous at x 0 . The following theorem follows directly from the definition of continuity, Theorem 3.1.2 and Theorem 3.3.2 and we leave its proof as an exercise. Theorem 3.3.3 Let f : D → R and let x 0 ∈ D . Then f is continuous at x 0 if and only if for any sequence { x k } in D that converges to x 0 , the sequence { f ( x k ) } converges to f ( x 0 ) . The proofs of the next two theorems are straightforward using Theorem 3.3.3 . Theorem 3.3.4 Let f , g : D → R and let x 0 ∈ D . Suppose f and g are continuous at x 0 . Then (a) f + g and f g are continuous at x 0 . (b) c f is continuous at x 0 for any constant c . (c) If g ( x 0 ) 6 = 0, then f g (defined on e D = { x ∈ D : g ( x ) 6 = 0 } ) is continuous at x 0 . Proof: We prove (a) and leave the other parts as an exercise. We will use Theorem 3.3.3 . Let { x k } be a sequence in D that converges to x 0 . Since f and g are continuous at x 0 , by Theorem 3.3.3 we obtain that { f ( x k ) } converges to f ( x 0 ) and { g ( x k ) } converges to g ( x 0 ) . By Theorem 2.2.1 (a) , we get that { f ( x k )+ g ( x k ) } converges to f ( x 0 )+ g ( x 0 ) . Therefore, lim k → ∞ ( f + g )( x k ) = lim k → ∞ f ( x k )+ g ( x k ) = f ( x 0 )+ g ( x 0 ) = ( f + g )( x 0 ) . Since { x k } was arbitrary, using Theorem 3.3.3 again we conclude f + g is continuous at x 0 . Theorem 3.3.5 Let f : D → R and let g : E → R with f ( D ) ⊂ E . If f is continuous at x 0 and g is continuous at f ( x 0 ) , then g ◦ f is continuous at x 0 .