107 Theorem 4.4.2 Let a,b ∈ R, a < b, and x0 ∈ (a,b). Let f ,g: (a,b) \{x0}→ R be 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), x ≠ x0, f ′ (x) (iii) limx →x0 = ℓ, for some ℓ ∈ R. g ′ (x) Then f (x) lim = ℓ. x→x0 g(x) 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). f (x) f (x) We will show that lim + = ℓ. The proof that lim − = ℓ is completely analogous. x →x x→x 0 g(x) 0 g(x) Fix any ε > 0. We need to fnd δ0 > 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) f ′ (x) ε ≤ K and − ℓ < (4.9) g ′ (x) g ′ (x) 2 whenever x ∈ (x0 − δ1,x0 + δ1) ∩ (a,b), x ̸ = x0. Fix α ∈ (x0,x0 + δ1) ∩ (a,b) (in particular, α > x0). Since limx →x0 f (x) = ∞, we can fnd 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, g(α) f (x) f (x) − f (α) 1 − g(x) = . g(x) g(x) − g(α) f (α) 1− f (x) Now, defne g(α) 1 − g(x) Hα (x)= for x ∈ (x0,x0 + δ2). f (α) 1− f (x) Since limx →x0 f (x)= limx →x0 g(x)= ∞, we have that lim + Hα (x)= 1. Thus, there exists a positive x →x0 γ < δ2 such that ε |Hα (x) − 1| < whenever x ∈ (x0,x0 + γ). 2K 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 such c we get f (x) f ′ (c) = Hα (x). g(x) g ′ (c)
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