109 ■ Example 4.4.6 Consider the limit 1 lim . x→∞ x( π 2 − arctanx) 1/x Writing the quotient in the form π , we can apply Theorem 4.4.3. We now compute the limit 2 −arctanx of the quotient of the derivatives 2 −(1/x2) x + 1 lim = lim = 1. x2 x→∞ 1 x→∞ − x2+1 In view of Theorem 4.4.3 the desired limit is also 1. The following theorem can be proved following the method in the proof of Theorem 4.4.2. Theorem 4.4.4 Let f and g be differentiable on (a,∞). Suppose that: (i) g ′ (x) ̸ = 0 for all x ∈ (a,∞), (ii) limx →∞ f (x)= limx →∞ g(x)= ∞, f ′ (x) (iii) limx →∞ = ℓ, for some ℓ ∈ R. g ′ (x) Then f (x) lim = ℓ. x→∞ g(x) ■ Example 4.4.7 Consider the limit lnx lim . x→∞ x Clearly the functions f (x)= lnx and g(x)= x satisfy the conditions of Theorem 4.4.4. We have f ′ (x) 1/x lim = lim = 0 x→∞ g ′ (x) x→∞ 1 lnx It follows from Theorem 4.4.4 that limx →∞ = 0. x Exercises 4.4.1 Use L’Hôpital’s rule to fnd the following limits (you may assume known all the relevant derivatives from calculus): 3 − 4 x x (a) lim . x→−2 3x2 + 5x − 2 ex − e−x (b) lim . x→0 sinxcosx
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