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 everyk,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 everyk ≥K. For anyδ >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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