105 (iii) limx→x0 f ′(x) g′(x) =ℓ, for some ℓ ∈R. Then lim x→x0 f (x) g(x) =ℓ. (4.9) Proof: Let {xn} be a sequence in [a,b] that converges to x0 and such that xn̸ =x0 for every n. By Theorem 4.2.4, for each n, there exists a sequence {cn}, withcn between xn andx0, such that [ f (xn)−f (x0)]g′(cn) = [g(xn)−g(x0)] f ′(cn). Since f (x0) =g(x0) =0, andg′(cn)̸ =0 for sufficiently large n, we have f (xn) g(xn) = f ′(cn) g′(cn) . Under the assumptions that g′(x)̸ =0 for x near x0 and g(x0) =0, we also have g(xn)̸ =0 for sufficiently large n. By the squeeze theorem (Theorem 2.1.3), {cn}converges to x0. Thus, lim k→∞ f (xn) g(xn) =lim n→∞ f ′(cn) g′(cn) = lim x→x0 f ′(x) g′(x) =ℓ. Therefore, (4.9) follows from Theorem 3.1.2. □ ■ Example 4.4.1 We will use Theorem 4.4.1 to show that lim x→0 2x+sinx x2 +3x =1. 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) g(x) is well defined on[a,b] \ {x0}. Moreover, taking δ =7/3, we get g′(x) =2x+3̸=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 lim x→x0 f ′(x) g′(x) =lim x→0 2+cosx 2x+3 = limx→02+limx→0cosx limx→0(2x+3) = 2+1 3 =1. It now follows from Theorem 4.4.1 that limx→0 2x+sinx x2 +3x =1 as we wanted to show. ■ Example 4.4.2 We will apply L’Hôpital’s rule to determine the limit lim x→1 3x3 −2x2 +4x−5 4x4 −2x−2 . Here f (x) =3x3 −2x2 +4x−5, g(x) =4x4 −2x−2 andx 0 =1. Thus f (1) =g(1) =0. Moreover, f ′(x) =9x2 −4x+4 and g′(x) =16x3 −2. Since g′(1) =14̸ =0 and g′ is continuous we have g′(x)̸ =0 for x near 1. Now, lim x→x0 f ′(x) g′(x) =lim x→1 9x2 −4x+4 16x3 −2 = 9 14 . Thus, the desired limit is 9 14 as well.
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