128 5.3 Limit Superior and Limit Inferior of Functions Remark 5.3.2 Let f : D→Rand let x0 be a limit point of D. Suppose limsupx →x0 f (x) is a real number. Define A={ℓ ∈R: ∃{xk} ⊂D,xk̸ =x0 for every k,xk →x0, f (xk) →ℓ}. Then the previous corollary shows that A̸=0/ and limsupx →x0 f (x) =maxA. Theorem 5.3.3 Let f : D→Rand let x0 be a limit point of D. Then limsup x→x0 f (x) =∞ if and only if there exists a sequence {xk}inDsuch that {xk}converges tox0, xk̸ =x0 for everyk, and limk→∞ f (xk) =∞. Proof: Suppose limsupx →x0 f (x) =∞. Then inf δ>0 g(δ) =∞, where gis the extended real-valued function defined in (5.2). Thus, g(δ) =∞for everyδ >0. Let δk = 1 k for k ∈N. Since g(δk) = sup x∈B0(x0;δk)∩D f (x) =∞, there exists xk ∈B0(x0;δk)∩Dsuch that f (xk) >k. Therefore, limk→∞ f (xk) =∞. Let us prove the converse. Since limk→∞ f (xk) =∞, for every M∈R, there exists K∈Nsuch that f (xk) ≥Mfor every k ≥K. For any δ >0, we have xk ∈B0(x0;δ)∩D whenever k is sufficiently large. Thus, g(δ) = sup x∈B0(x0;δ)∩D f (x) ≥M. This implies g(δ) =∞, and hence limsupx →x0 f (x) =∞. □ Theorem 5.3.4 Let f : D→Rand let x0 be a limit point of D. Then limsup x→x0 f (x) =−∞ if and only if for any sequence {xk} in Dsuch that {xk} converges to x0, xk̸ =x0 for every k, it follows that limk→∞ f (xk) =−∞. The latter is equivalent to limx→x0 f (x) =−∞.
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