Chapter 4 Limits and Continuous Functions 4.1 Limits of real valued functions Definition 4.1 Let a ∈ R. (c, d) such that c < a < d is called a neighbourhood of a, and (c, d) \ {a} is called a deleted neighbourhood of a. Definition 4.2 ’ f (x) → L as x → a’ written lim f (x) = L means: x→a ∀ ε > 0 ∃ δ > 0 ∀ x ∈ (a − δ, a + δ) \ {a}, f (x) ∈ (L − ε, L + ε), i. e., ∀ ε > 0 ∃ δ > 0 (0 < |x − a| < δ ⇒ | f (x) − L| < ε). 1 Example 4.1 Prove, from first principles, that x2 → 4 as x → 2. Solution. Let ε > 0. We must show that there is δ > 0 such that 0 < |x − 2| < δ implies |x2 − 4| < ε. Since we want to use |x − 2| < δ to get |x2 − 4| < ε, we try to factor out |x − 2| from |x2 − 4|. Thus |x2 − 4| = |x − 2| |x + 2| = |x − 2| |x − 2 + 4| ≤ |x − 2| (|x − 2| + 4). It is often convenient to make an initial restriction on δ, like δ ≤ 1. Then we obtain from |x − 2| < δ that |x2 − 4| ≤ |x − 2| (1 + 4) = 5|x − 2|. If we now put ε δ = min 1, > 0, 5 then we can conclude (note that we already used δ ≤ 1) for 0 < |x − 2| < δ that ε |x2 − 4| ≤ 5|x − 2| < 5δ ≤ 5 = ε. 5 2 Theorem 4.1 If f (x) → L as x → a, then L is unique. Proof. Let f (x) → L and f (x) → M as x → a. Let’s show that L = M . So let ε > 0. Then there are δ1 > 0 and δ2 > 0 such that ε ε and 0 < |x−a| < δ2 ⇒ | f (x)−M| < . 2 2 For δ = min{δ1, δ2} and 0 < |x − a| < δ, it follows that ε ε |L − M| = |(L − f (x)) − ( f (x) − M)| ≤ | f (x) − L| + | f (x) − M| < + = ε. 2 2 Hence L = M by Lemma 2.1. 0 < |x−a| < δ1 ⇒ | f (x)−L| < 3