127 Let us now prove the converse. Suppose (i) and (ii) are satisfied. Fix any ε >0 and let δ >0 satisfy (i). Then g(δ) = sup x∈B0(x0;δ)∩D f (x) ≤ℓ+ε. This implies limsup x→x0 f (x) =inf δ>0 g(δ) ≤ℓ+ε. Since ε is arbitrary, we get limsup x→x0 f (x) ≤ℓ. Again, take any ε >0. Givenδ >0, let ˆ x be as in (ii). Then ℓ−ε <f (ˆ x) ≤ sup x∈B0(x0;δ)∩D f (x) =g(δ). This implies ℓ−ε ≤inf δ>0 g(δ) =limsup x→x0 f (x). It follows that ℓ ≤limsupx →x0 f (x). Therefore, ℓ =limsupx →x0 f (x). □ Corollary 5.3.2 Suppose ℓ =limsupx →x0 f (x), where ℓ is a real number. Then there exists a sequence {xk}inDsuch that {xk}converges to x0, xk̸ =x0 for every k, and lim k→∞ f (xk) =ℓ. Moreover, if {yk}is a sequence inDthat converges tox0, yk̸ =x0 for everyk, and limk→∞ f (yk) =ℓ′, thenℓ′ ≤ℓ. Proof: For each k ∈N, let εk = 1 k. By Condition (i) of Theorem 5.3.1, there exists δk >0 such that f (x) < ℓ+εk for all x ∈B0(x0;δk)∩D. (5.3) Let δ′ k =min{δk, 1 k}. Thenδ′ k ≤δk and limk→∞δ′ k =0. From Condition (ii) of Theorem 5.3.1, there exists xk ∈B0(x0;δ′ k)∩Dsuch that ℓ−εk <f (xk). Moreover, f (xk) < ℓ+εk by (5.3). Therefore, {xk}is a sequence that satisfies the conclusion of the corollary. Now let {yk}be a sequence inDthat converges tox0, yk̸ =x0 for everyk, and limk→∞ f (yk) =ℓ′. For any ε >0, let δ >0 be as in Condition (i) of Theorem 5.3.1. Since yk ∈B0(x0;δ)∩Dwhen k is sufficiently large, we have f (yk) < ℓ+ε for such k. This implies ℓ′ ≤ℓ+ε. It follows that ℓ′ ≤ℓ. □
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