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) g(x) −ℓ = f ′(c) g′(c) Hα(x)−ℓ = f ′(c) g′(c) (Hα(x)−1)+ f ′(c) g′(c) −ℓ ≤ f ′(c) g′(c) | Hα(x)−1|+ f ′(c) g′(c) −ℓ <K ε 2K + ε 2 =ε. Settingδ0 =γ completes the proof. □ ■ Example 4.4.5 Consider the limit lim x→0 lnx2 1+ 1 3√x2 . Here f (x)=lnx2, g(x)=1+ 1 3√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 x̸ =0. Moreover, limx →x0 f (x)=limx →x0 g(x)=∞. We analyze the quotient of the derivatives. We have lim x→0 2/x −2 3 1 3√x5 =lim x→0 (−3) 3√x5 x =lim x→0 (−3) 3√x2 =0. It now follows from Theorem 4.4.2 that lim x→0 lnx2 1+ 1 3√x2 =0. 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 modified to include the case whenx0 is an endpoint of the domain of the functions f andg. The following theorem can be proved following the method in the proof of Theorem 4.4.1. Theorem 4.4.3 Let f andg 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, (iii) limx →∞ f ′(x) g′(x) =ℓ, for some ℓ ∈R. Then lim x→∞ f (x) g(x) =ℓ.
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