129 Proof: Suppose limsupx →x0 f (x)= −∞. Take any sequence {xk} in D such that {xk} converges to x0 and xk ̸ = x0 for every k. Fix any M ∈ R. By the defnition of limit superior, there exists δ > 0 such that g(δ )= sup f (x) < M, x∈B0(x0;δ )∩D which implies that f (x) < M for all x ∈ B0(x0;δ ) ∩ D. Since {xk}⊂ D converges to x0 and xk ̸ = x0 for every k, there exists K ∈ N such that xk ∈ B0(x0;δ ) ∩ D for all k ≥ K. This implies f (xk) < M for all k ≥ K. Therefore, limk →∞ f (xk)= −∞. To prove the converse, we assume that for any sequence {xk} in D such that {xk} converges to x0, xk ̸ = x0 for every k, it follows that limk →∞ f (xk)= −∞. Then it is not hard to show that limx →x0 f (x)= −∞. Thus, for any M ∈ R, there exists δ > 0 such that f (x) ≤ M for all x ∈ B0(x0;δ ), which implies that g(δ )= sup f (x) ≤ M. x∈B0(x0;δ )∩D Then limsupx →x0 f (x)= infδ >0 g(δ ) ≤ M. Therefore, limsupx →x0 f (x)= −∞ since M is arbitrary. □ Following the same arguments, we can prove similar results for inferior limits of functions. Theorem 5.3.5 Let f : D → R, let x0 be a limit point of D, and let ℓ be a real number. Then ℓ = liminfx →x0 f (x) if and only if the following two conditions hold: (i) For every ε > 0, there exists δ > 0 such that ℓ − ε < f (x) for all x ∈ B0(x0;δ ) ∩ D; (ii) For every ε > 0 and for every δ > 0, there exists x ∈ B0(x0;δ ) ∩ D such that f (x) <ℓ + ε. Corollary 5.3.6 Suppose ℓ = liminfx →x0 f (x), where ℓ is a real number. Then there exists a sequence {xk} in D such that xk converges to x0, xk ̸ = x0 for every k, and lim f (xk)= ℓ. k→∞ Moreover, if {yk} is a sequence in D that converges to x0, yk ̸ = x0 for every k, and limk →∞ f (yk)= ℓ ′ , then ℓ ′ ≥ ℓ.
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