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 → R and let x0 be a limit point of D. Then the following are equivalent: (i) lim f (x)= ℓ. x→x0 (ii) lim f (xn)= ℓ for every sequence {xn} in D \{x0} such that xn → x0. n→∞ Proof: We frst prove that (i) implies (ii). Suppose (i) holds. Let {xn} be a sequence in D with xn ̸ = x0 for every n and 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 ∈ N such 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/n where n is a positive integer. Thus, for every n ∈ N, there exists xn ∈ D with 1 0 < |x n − x0| < and | f (xn) − ℓ|≥ ε0. By the squeeze theorem (Theorem 2.1.3), the sequence {xn} n converges to x0. Moreover, xn ̸ = 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 → R and 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} in D\{x0} such that {xn} converges to x0, and { f (xn)} does not converge. ■ Example 3.1.6 Consider f : R\{0}→ R given by f (x)= cos(1/x). We will prove that limx →0 f (x) 1 does not exist. For that, consider the sequence {x n} given by xn = nπ for n ∈ N. Then limn →∞ xn = 0 and xn ̸ = 0 for all n ∈ N, that is, {xn} is in R \{0}. We have lim f (xn)= lim cos(1/xn)= lim cos(nπ)=(−1) n . n→∞ n→∞ 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 → R and 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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