64 3.1 Limits of Functions The following theorem will let us apply our earlier results on limits of sequences to obtain new results on limits of functions. Theorem 3.1.2 — Sequential Characterization of Limits. Let f : D→Rand let x0 be a limit point of D. Then the following are equivalent: (i) lim x→x0 f (x) =ℓ. (ii) lim n→∞ f (xn) =ℓ for every sequence {xn}inD\ {x0}such that xn →x0. Proof: We first prove that (i) implies (ii). Suppose (i) holds. Let {xn}be a sequence inDwithxn̸ =x0 for every nand such that {xn}converges to x0. We now proceed to show that limn→∞ f (xn) =ℓ. Let ε >0. From the assumption (i) we know there exists δ >0 such that | f (x)−ℓ| <ε whenever x ∈D and 0<|x−x0| <δ. Since xn →x0, there exists N∈Nsuch that 0<|xn −x0| <δ for all n≥N. For such n, we have | f (xn)−ℓ| <ε. This shows that limn→∞ f (xn) =ℓ and, thus, (ii) follows. We now prove that (ii) implies (i). We proceed by contradiction. We assume that (ii) is true but (i) is false. Since (i) is false, there exists ε0 >0 such that for every δ >0, there exists x ∈D with 0<|x−x0| <δ and| f (x)−ℓ| ≥ε0. We will use this fact about ε0 with different choices of δ, namely, with δ =1/nwhere nis a positive integer. Thus, for every n∈N, there exists xn ∈Dwith 0<|xn −x0| < 1 n and | f (xn)−ℓ| ≥ε0. By the squeeze theorem (Theorem 2.1.3), the sequence {xn} converges to x0. Moreover, xn̸ =x0 for every n. On the other hand, the inequality with ε0 shows that that the sequence {f (xn)}does not converge to ℓ. This contradicts (ii). It follows that (ii) implies (i) and the proof is complete. □ A useful application of Theorem 3.1.2 is in proving that the limit of a function does not exist at some point. The following corollaries illustrate two approaches. The proofs follow immediately from the theorem and are left as exercises. Corollary 3.1.3 Let f : D→Rand let x0 be a limit point of D. Then f does not have a limit at x0 if and only if there exists a sequence {xn}inD\ {x0}such that {xn}converges tox0, and{f (xn)} does not converge. ■Example 3.1.6 Consider f : R\{0}→Rgiven by f (x) =cos(1/x). We will prove that limx→0 f (x) does not exist. For that, consider the sequence {xn}given byxn = 1 nπ for n∈N. Then limn→∞xn =0 andxn̸ =0 for all n∈N, that is, {xn}is inR\ {0}. We have lim n→∞ f (xn) =lim n→∞ cos(1/xn) =lim n→∞ cos(nπ) = (−1)n. Since this limit does not exist, applying corollary 3.1.3 we conclude that limx→0 f (x) does not exist. Corollary 3.1.4 Let f : D→Rand let x0 be a limit point of D. If there exist two sequences {xn} and {yn} in D\ {x0} such that both sequences {xn} and {yn} converge to x0 and limn→∞ f (xn)̸ = limn→∞ f (yn), then f does not have a limit at x0.
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