Introduction to Mathematical Analysis I - 3rd Edition

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 every k, 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 any M∈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, limsupx →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 every δ >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 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.

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