65 ■ Example 3.1.7 Consider the Dirichlet function f : R→Rgiven by f (x) =(1, if x ∈Q; 0, if x ∈Qc. We will show that the limx→x0 f (x) does not exist for any x0 ∈R. For that, consider x0 ∈Rand choose two sequences {rn}, {sn} converging to x0 such that rn ∈Qand sn̸ ∈Qfor 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 inR. Theorem 3.1.5 — Comparison Theorem for Functions. Let f,g: D→Rand 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,x̸ =x0. Then ℓ1 ≤ℓ2. Proof: Let {xn}be a sequence in(x0 −δ,x0 +δ)∩Dthat converges to x0 and xn̸ =x0 for all n. By Theorem 3.1.2, lim n→∞ f (xn) =ℓ1 and lim n→∞ g(xn) =ℓ2. Since f (xn) ≤g(xn) for all n∈N, applying Theorem 2.1.2, we obtainℓ1 ≤ℓ2. □ Theorem 3.1.6 Let f,g: D→Rand 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,x̸ =x0. Proof: Choose ε >0 such that ℓ1 +ε < ℓ2 −ε (equivalently, such that ε < ℓ2−ℓ1 2 ). Then there exists δ >0 such that ℓ1 −ε <f (x) < ℓ1 +ε andℓ2 −ε <g(x) < ℓ2 +ε for all x ∈(x0 −δ,x0 +δ)∩D,x̸ =x0. Thus, f (x) < ℓ1 +ε < ℓ2 −ε <g(x) for all x ∈(x0 −δ,x0 +δ)∩D,x̸ =x0. The proof is now complete. □ Theorem 3.1.7 — Squeeze Theorem for Functions. Let f,g,h: D→Rand 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,x̸ =x0,
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