128 5.3 Limit Superior and Limit Inferior of Functions Remark 5.3.2 Let f : D → R and let x0 be a limit point of D. Suppose limsupx →x0 f (x) is a real number. Defne A = {ℓ ∈ R : ∃{xk}⊂ D,xk ≠ x0 for every k,xk → x0, f (xk) → ℓ}. Then the previous corollary shows that A ̸ = 0/ and limsupx →x0 f (x)= maxA. Theorem 5.3.3 Let f : D → R and let x0 be a limit point of D. Then limsup f (x)= ∞ x→x0 if and only if there exists a sequence {xk} in D such that {xk} converges to x0, xk ̸ = x0 for every k, and limk →∞ f (xk)= ∞. Proof: Suppose limsupx →x0 f (x)= ∞. Then inf g(δ )= ∞, δ >0 where g is the extended real-valued function defned in (5.2). Thus, g(δ )= ∞ for every δ > 0. Let δk = 1 k for k ∈ N. Since g(δk)= sup f (x)= ∞, x∈B0(x0;δk)∩D there exists xk ∈ B0(x0;δk) ∩ D such that f (xk) > k. Therefore, limk →∞ f (xk)= ∞. Let us prove the converse. Since limk →∞ f (xk)= ∞, for every M ∈ R, there exists K ∈ N such that f (xk) ≥ M for every k ≥ K. For any δ > 0, we have xk ∈ B0(x0;δ ) ∩ D whenever k is suffciently large. Thus, g(δ )= sup f (x) ≥ M. x∈B0(x0;δ )∩D This implies g(δ )= ∞, and hence limsupx →x0 f (x)= ∞. □ Theorem 5.3.4 Let f : D → R and let x0 be a limit point of D. Then limsup f (x)= −∞ x→x0 if and only if for any sequence {xk} in D such that {xk} converges to x0, xk ̸ = x0 for every k, it follows that limk →∞ f (xk)= −∞. The latter is equivalent to limx →x0 f (x)= −∞.
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