107 Theorem 4.4.2 Let a,b∈R, a<b, and x0 ∈(a,b). Let f,g: (a,b)\ {x0} →Rbe differentiable on (a,b)\ {x0}. Suppose that: (i) limx→x0 f (x) =limx→x0 g(x) =∞, (ii) there exists δ >0 such that g′(x)̸ =0 for all x ∈(x0 −δ,x0 +δ)∩(a,b), x̸ =x0, (iii) limx→x0 f ′(x) g′(x) =ℓ, for some ℓ ∈R. Then lim x→x0 f (x) g(x) =ℓ. Proof: Since limx→x0 f (x) =limx→x0 g(x) =∞, choosing a smaller positive δ if necessary, we can assume that f (x)̸ =0 and g(x)̸ =0 for all x ∈(x0 −δ,x0 +δ)∩(a,b). We will show that limx →x+ 0 f (x) g(x) =ℓ. The proof that limx →x− 0 f (x) g(x) =ℓ is completely analogous. Fix anyε >0. We need to findδ0 >0 such that | f (x)/g(x)−ℓ| <ε whenever x∈(x0,x0+δ0)∩ (a,b). From (iii), one can choose K>0 and a positive δ1 <δ such that f ′(x) g′(x) ≤ Kand f ′(x) g′(x) −ℓ < ε 2 (4.10) whenever x ∈(x0 −δ1,x0 +δ1)∩(a,b), x̸ =x0. Fix α ∈(x0,x0 +δ1)∩(a,b) (in particular, α >x0). Since limx→x0 f (x) =∞, we can find δ2 >0 such that δ2 <min{δ1,α−x0}and f (x)̸ =f (α) for x ∈(x0,x0+δ2)∩(a,b) = (x0,x0+δ2). Moreover, for such x, since g′(z)̸ =0 if x <z <α, Rolle’s theorem (Theorem 4.2.2) guarantees that g(x)̸ =g(α). Therefore, for all x ∈(x0,x0 +δ2) we can write, f (x) g(x) = f (x)−f (α) g(x)−g(α) 1− g(α) g(x) 1− f (α) f (x) . Now, define Hα(x) = 1− g(α) g(x) 1− f (α) f (x) for x ∈(x0,x0 +δ2). Since limx→x0 f (x) =limx→x0 g(x) =∞, we have that limx →x+ 0 Hα(x) =1. Thus, there exists a positive γ <δ2 such that |Hα(x)−1| < ε 2K whenever x ∈(x0,x0 +γ). For any x ∈(x0,x0 +γ), applying Theorem 4.2.4 on the interval [x,α], we can write [ f (x) − f (α)]g′(c) = [g(x)−g(α)] f ′(c) for some c ∈(x,α) (note that, in particular, c ∈(x0 −δ1,x0 + δ1)∩(a,b)). For suchc we get f (x) g(x) = f ′(c) g′(c) Hα(x).
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