65 ■ Example 3.1.7 Consider the Dirichlet function f : R → R given by (1, if x ∈ Q; f (x)= 0, if x ∈ Qc . We will show that the limx →x0 f (x) does not exist for any x0 ∈ R. For that, consider x0 ∈ R and choose two sequences {rn}, {sn} converging to x0 such that rn ∈ Q and sn ̸ ∈ Q for all n ∈ N (see Exercises 2.1.8 and 2.1.9). Since f (rn)= 1 for all n ∈ N, the sequence { f (rn)} converges to 1 and since f (sn)= 0 for all n ∈ N, the sequence { f (sn)} converges to 0. Applying corollary 3.1.4 we conclude that limx →x0 f (x) does not exist. Since x0 was an arbitrary real number, the Dirichlet function does not have a limit at any point in R. Theorem 3.1.5 — Comparison Theorem for Functions. Let f ,g: D → R and let x0 be a limit point of D. Suppose that (i) limx →x0 f (x)= ℓ1, limx →x0 g(x)= ℓ2, (ii) there exists δ > 0 such that f (x) ≤ g(x) for all x ∈ (x0 − δ ,x0 + δ ) ∩ D,x ̸ = x0. Then ℓ1 ≤ ℓ2. Proof: Let {xn} be a sequence in (x0 − δ ,x0 + δ ) ∩ D that converges to x0 and xn ̸ = x0 for all n. By Theorem 3.1.2, lim f (xn)= ℓ1 and lim g(xn)= ℓ2. n→∞ n→∞ Since f (xn) ≤ g(xn) for all n ∈ N, applying Theorem 2.1.2, we obtain ℓ1 ≤ ℓ2. □ Theorem 3.1.6 Let f ,g: D → R and let x0 be a limit point of D. Suppose that (i) limx →x0 f (x)= ℓ1, limx →x0 g(x)= ℓ2, (ii) ℓ1 <ℓ2. Then there exists δ > 0 such that f (x) < g(x) for all x ∈ (x0 − δ ,x0 + δ ) ∩ D,x ̸ = x0. ℓ2−ℓ1 Proof: Choose ε > 0 such that ℓ1 + ε <ℓ2 − ε (equivalently, such that ε < ). Then there exists 2 δ > 0 such that ℓ1 − ε < f (x) <ℓ1 + ε and ℓ2 − ε < g(x) <ℓ2 + ε for all x ∈ (x0 − δ ,x0 + δ ) ∩ D,x ̸ = x0. Thus, f (x) <ℓ1 + ε <ℓ2 − ε < g(x) for all x ∈ (x0 − δ ,x0 + δ ) ∩ D,x ̸ = x0. The proof is now complete. □ Theorem 3.1.7 — Squeeze Theorem for Functions. Let f,g,h: D → R and let x0 be a limit point of D. Suppose that (i) there exists δ > 0 such that f (x) ≤ g(x) ≤ h(x) for all x ∈ (x0 − δ ,x0 + δ ) ∩ D,x ̸ = x0,
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