Introduction to Mathematical Analysis I - Second Edition

64 3.1 LIMITS OF FUNCTIONS Example 3.1.3 Let f : R → R be given by f ( x ) = x 2 . We show that lim x → 2 f ( x ) = 4. First note that | f ( x ) − 4 | = | x 2 − 4 | = | ( x − 2 )( x + 2 ) | = | x − 2 || x + 2 | . Since the domain is all of R the expression | x + 2 | is not bounded and we cannot proceed as in Example 3.1.2 . However, we are interested only in values of x close to 2 and, thus, we impose the condition δ ≤ 1. If | x − 2 | < 1, then − 1 < x − 2 < 1 and, so, 1 < x < 3. It follows, for such x , that | x | < 3 and, hence | x | + 2 < 5. Now, given ε > 0 we choose δ = min { 1 , ε 5 } . Then, whenver | x − 2 | < δ we get | f ( x ) − 4 | = | x − 2 | x + 2 | ≤ | x − 2 | ( | x | + 2 ) < δ 5 ≤ ε . Example 3.1.4 Let f : R → R be given by f ( x ) = 3 x − 5 x 2 + 3 . We prove that lim x → 1 f ( x ) = − 1 2 . First we look at the expression | f ( x ) − ( − 1 2 ) | and try to identify a factor | x − 1 | (because here ¯ x = 1). f ( x ) − − 1 2 = 3 x − 5 x 2 + 3 + 1 2 = 6 x − 10 + x 2 + 3 x 2 + 3 = | x − 1 || x + 7 | 2 ( x 2 + 3 ) ≤ 1 6 | x − 1 || x + 7 | . Proceeding as in the previous example, if | x − 1 | < 1 we get − 1 < x − 1 < 1 and, so, 0 < x < 2. Thus | x | < 2 and | x + 7 | ≤ | x | + 2 < 9. Now, given ε > 0, we choose δ = min { 1 , 2 3 ε } . It follows that if | x − 1 | < δ we get f ( x ) − ( − 1 2 ) ≤ | x + 7 | 6 | x − 1 | < 9 6 δ ≤ ε . The following theorem will let us apply our earlier results on limits of sequences to obtain new results on limits of functions. Theorem 3.1.2 — Sequential Characterization of Limits. Let f : D → R and let ¯ x be a limit point of D . Then lim x → ¯ x f ( x ) = ` (3.1) if and only if lim n → ∞ f ( x n ) = ` (3.2) for every sequence { x n } in D such that x n 6 = ¯ x for every n and { x n } converges to ¯ x . Proof: Suppose ( 3.1 ) holds. Let { x n } be a sequence in D with x n 6 = ¯ x for every n and such that { x n } converges to ¯ x . Given any ε > 0, there exists δ > 0 such that | f ( x ) − ` | < ε whenever x ∈ D and 0 < | x − ¯ x | < δ . Then there exists N ∈ N with 0 < | x n − ¯ x | < δ for all n ≥ N . For such n , we have | f ( x n ) − ` | < ε . This implies ( 3.2 ) . Conversely, suppose ( 3.1 ) is false. Then there exists ε 0 > 0 such that for every δ > 0, there exists x ∈ D with 0 < | x − ¯ x | < δ and | f ( x ) − ` | ≥ ε 0 . Thus, for every n ∈ N , there exists x n ∈ D with 0 < | x n − ¯ x | < 1 n and | f ( x n ) − ` | ≥ ε 0 . By the squeeze theorem (Theorem 2.1.6 ) , the sequence { x n } converges to ¯ x . Moreover, x n 6 = ¯ x for every n . This shows that ( 3.2 ) is false. It follows that ( 3.2 ) implies ( 3.1 ) and the proof is complete.