129 Proof: Suppose limsupx →x0 f (x)=−∞. Take any sequence {xk}inDsuch that {xk}converges to x0 and xk̸ =x0 for every k. Fix any M∈R. By the definition of limit superior, there exists δ >0 such that g(δ)= sup x∈B0(x0;δ)∩D f (x) <M, which implies that f (x) <Mfor all x ∈B0(x0;δ)∩D. Since {xk}⊂Dconverges tox0 andxk̸ =x0 for everyk, there exists K∈Nsuch that xk ∈B0(x0;δ)∩Dfor all k ≥K. This implies f (xk) <Mfor all k ≥K. Therefore, limk →∞ f (xk)=−∞. To prove the converse, we assume that for any sequence {xk} in Dsuch that {xk} converges to x0, xk̸ =x0 for every k, it follows that limk →∞ f (xk)=−∞. Then it is not hard to show that limx →x0 f (x)=−∞. Thus, for anyM∈R, there exists δ >0 such that f (x) ≤Mfor all x ∈B0(x0;δ), which implies that g(δ)= sup x∈B0(x0;δ)∩D f (x) ≤M. Then limsupx →x0 f (x)=infδ>0g(δ) ≤M. Therefore, lim supx →x0 f (x)=−∞since Mis 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 everyε >0, there exists δ >0 such that ℓ−ε <f (x) for all x ∈B0(x0;δ)∩D; (ii) For everyε >0 and for everyδ >0, there exists x ∈B0(x0;δ)∩Dsuch that f (x) <ℓ+ε. Corollary 5.3.6 Supposeℓ=liminfx →x0 f (x),whereℓ is a real number. Then there exists a sequence {xk} inDsuch that xk converges tox0, xk̸ =x0 for everyk, 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.
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