105 f ′ (x) (iii) limx→x0 = ℓ, for some ℓ ∈ R. g ′ (x) Then f (x) lim = ℓ. (4.8) x→x0 g(x) Proof: Let {xn} be a sequence in [a,b] that converges to x0 and such that xn ̸ = x0 for every n. By Theorem 4.2.4, for each n, there exists a sequence {cn}, with cn between xn and x0, such that [ f (xn) − f (x0)]g ′ (cn)=[g(xn) − g(x0)] f ′ (cn). Since f (x0)= g(x0)= 0, and g ′ (cn) ̸ = 0 for suffciently large n, we have f (xn) f ′ (cn) = . g(xn) g ′ (cn) Under the assumptions that g ′ (x)≠ 0 for x near x0 and g(x0)= 0, we also have g(xn)≠ 0 for suffciently large n. By the squeeze theorem (Theorem 2.1.3), {cn} converges to x0. Thus, f (xn) f ′ (cn) f ′ (x) lim = lim = lim = ℓ. k→∞ g(xn) n→∞ g ′ (cn) x→x0 g ′ (x) Therefore, (4.8) follows from Theorem 3.1.2. □ ■ Example 4.4.1 We will use Theorem 4.4.1 to show that 2x + sinx lim = 1. x→0 x2 + 3x First we observe that the conditions of Theorem 4.4.1 hold. Here f (x)= 2x + sinx, g(x)= x2 + 3x, and x0 = 0. We may take [a,b] = [−1,1], for example, so that f and g are continuous on [a,b] and differentiable on (a,b) and, furthermore, f (x) is well defned on [a,b] \{x0}. Moreover, taking g(x) δ = 7/3, we get g ′ (x)= 2x + 3 ≠ 0 for x ∈ (x0 − δ ,x0 + δ ) ∩ [a,b]. Finally we calculate the limit of the quotient of derivatives using Theorem 3.2.1 to get f ′ (x) 2 + cosx limx →0 2+ limx →0 cosx 2+ 1 lim = lim = = = 1. x→x0 g ′ (x) x→0 2x + 3 limx →0(2x + 3) 3 2x + sinx It now follows from Theorem 4.4.1 that limx →0 = 1 as we wanted to show. x2 + 3x ■ Example 4.4.2 We will apply L’Hôpital’s rule to determine the limit 3 − 2x2 3x + 4x − 5 lim . x→1 4x4 − 2x − 2 3 − 2x2 Here f (x)= 3x + 4x − 5, g(x)= 4x4 − 2x − 2 and x 0 = 1. Thus f (1)= g(1)= 0. Moreover, f ′ (x)= 9x2 − 4x + 4 and g ′ (x)= 16x3 − 2. Since g ′ (1)= 14 ≠ 0 and g ′ is continuous we have g ′ (x) ̸ = 0 for x near 1. Now, f ′ (x) 9x2 − 4x + 4 9 lim = lim = . x→x0 g ′ (x) x→1 16x3 − 2 14 9 Thus, the desired limit is 14 as well.
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