108 4.4 L’Hôpital’s Rule Since c ∈ (x0 − δ1,x0 + δ1) ∩ (a,b), applying (4.9) we get that, for x ∈ (x0,x0 + γ)=(x0,x0 + γ) ∩ (a,b), f (x) f ′ (c) − ℓ = Hα (x) − ℓ g(x) g ′ (c) f ′ (c) f ′ (c) = (Hα (x) − 1)+ − ℓ g ′ (c) g ′ (c) f ′ (c) f ′ (c) ≤ |Hα (x) − 1| + − ℓ g ′ (c) g ′ (c) ε ε < K + = ε. 2K 2 Setting δ0 = γ completes the proof. □ ■ Example 4.4.5 Consider the limit lnx2 lim . x→0 1 + √3 1 x2 Here f (x)= lnx2, g(x)= 1+ √3 1 x2 , x0 = 0, and we may take as (a,b) any open interval containing 0. Clearly f and g satisfy the differentiability assumptions and g ′ (x)≠ 0 for all x ≠ 0. Moreover, limx →x0 f (x)= limx →x0 g(x)= ∞. We analyze the quotient of the derivatives. We have √ x5 √ 2/x 3 3 x2 = 0. lim = lim(−3) = lim(−3) x→0 −2 1 x→0 x x→0 3 √3 x5 It now follows from Theorem 4.4.2 that lnx2 lim = 0. 1 x→0 1 + √3 x2 Remark 4.4.1 The proofs of Theorem 4.4.1 and Theorem 4.4.2 show that the results in these theorems can be applied for left-hand and right-hand limits. Moreover, the results can also be modifed to include the case when x0 is an endpoint of the domain of the functions f and g. The following theorem can be proved following the method in the proof of Theorem 4.4.1. Theorem 4.4.3 Let f and g be differentiable on (a,∞). Suppose that: (i) g(x)≠ 0 and g ′ (x)≠ 0 for all x ∈ (a,∞), (ii) limx →∞ f (x)= limx →∞ g(x)= 0, f ′ (x) (iii) limx →∞ = ℓ, for some ℓ ∈ R. g ′ (x) Then f (x) lim = ℓ. x→∞ g(x)
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