L'Hopital's Rule: A Mathematical Theorem for Limits, Lecture notes of Pre-Calculus

Download L'Hopital's Rule: A Mathematical Theorem for Limits and more Lecture notes Pre-Calculus in PDF only on Docsity! 2017-18 MATH1010 Lecture 15: L’Hôpital’s Rule Charles Li 1 L’Hôpital’s Rule (Source: mooculus textbook) Derivatives allow us to take problems that were once difficult to solve and convert them to problems that are easier to solve. Let us consider l’Hôpital’s rule: Theorem 1.1 (L’Hôpital’s Rule). Let f(x) and g(x) be functions that are differentiable near a. If lim x→a f(x) = lim x→a g(x) = 0 or ±∞, and limx→a f ′(x) g′(x) exists, and g′(x) 6= 0 for all x near a, then lim x→a f(x) g(x) = lim x→a f ′(x) g′(x) .  Remark 1. L’Hôpital’s rule applies even when limx→a f(x) = ±∞ and limx→a g(x) = ±∞. 2. a can be +∞ and −∞. This theorem is somewhat difficult to prove, in part because it in- corporates so many different possibilities, we will prove the special case when a is finite. Proof. Here we proof the special type 0/0 and a finite. By Cauchy’s mean value theorem, there exists ξ between a and x such that f(x) g(x) = f(x)− f(a) g(x)− g(a) = f ′(ξ) g′(ξ) . When x→ a, xi (which depends on x) also tends to a. So lim x→a f(x) g(x) = lim ξ→a f ′(ξ) g′(ξ) . 1 Next we prove the special case∞/∞ and a finite (can be skipped). Suppose lim x→a f(x) =∞, lim x→a g(x) =∞. Let b be a number very closed to a, such that g′(x) is nonzero for x between a and b. Then by Cauchy’s mean value theorem, there exists ξ, between a and b, such that f(x)− f(b) g(x)− g(b) = f ′(ξ) g′(ξ) . Then we have f(x) g(x) = f ′(ξ) g′(ξ) · 1− g(b) g(x) 1− f(b) f(x) . Because lim ξ→0 f ′(ξ) g′(ξ) exists. Also lim x→∞ g(b) g(x) = 0 = lim x→∞ f(b) f(x) . Hence lim x→a 1− g(b) g(x) 1− f(b) f(x) = 1 1 = 1. So lim x→a f(x) g(x) = lim ξ→a f ′(ξ) g′(ξ) . L’Hôpital’s rule allows us to investigate limits of indeterminate form. Definition 1.1. 0/0 This refers to a limit of the form limx→a f(x) g(x) where f(x)→ 0 and g(x)→ 0 as x→ a. ∞/∞ This refers to a limit of the form limx→a f(x) g(x) where f(x)→∞ and g(x)→∞ as x→ a. 0 ·∞ This refers to a limit of the form limx→a (f(x) · g(x)) where f(x)→ 0 and g(x)→∞ as x→ a. ∞–∞ This refers to a limit of the form limx→a (f(x)− g(x)) where f(x)→∞ and g(x)→∞ as x→ a. 1∞ This refers to a limit of the form limx→a f(x)g(x) where f(x)→ 1 and g(x)→∞ as x→ a. 2 This situation is ripe for l’Hôpital’s Rule. Now f ′(x) = sec(x) tan(x) and g′(x) = sec2(x). L’Hôpital’s rule tells us that lim x→π/2+ sec(x) tan(x) = lim x→π/2+ sec(x) tan(x) sec2(x) = lim x→π/2+ sin(x) = 1.  Example 1.5 (∞/∞). Compute lim x→+∞ ln(ex + 1) ln(e2x + 1) .  Answer. Let f(x) = ln(ex + 1) and g(x) = ln(e2x + 1). Then limx→+∞ f(x) = +∞ and limx→+∞ g(x) = +∞. By l’Hôpital’s Rule lim x→+∞ f(x) g(x) = lim x→+∞ f ′(x) g′(x) lim x→+∞ ex ex+1 2e2x e2x+1 = lim x→+∞ e2x + 1 2ex(ex + 1) = lim x→+∞ 1 + e−2x 2(1 + e−x) = 1 2 .  Example 1.6 (0 ·∞). Compute lim x→0+ x lnx.  Answer. This doesn’t appear to be suitable for l’Hôpital’s Rule. As x approaches zero, ln x goes to −∞, so the product looks like (something very small) · (something very large and negative). This product could be anything—a careful analysis is required. Write x lnx = lnx x−1 . 5 Set f(x) = ln(x) and g(x) = x−1. Since both functions are differen- tiable near zero and lim x→0+ ln(x) = −∞ and lim x→0+ x−1 =∞, we may apply l’Hôpital’s rule. Write f ′(x) = x−1 and g′(x) = −x−2, so lim x→0+ x lnx = lim x→0+ lnx x−1 = lim x→0+ x−1 −x−2 = lim x→0+ −x = 0. One way to interpret this is that since limx→0+ x lnx = 0, the func- tion x approaches zero much faster than lnx approaches −∞.  Example 1.7. lim x→0+ x ln(1 + 3 x ).  Answer. x ln(1 + 3 x ) = ln(1 + 3 x ) 1 x . Let f(x) = ln(1 + 3 x ) and g(x) = 1 x . Also limx→0+ ln(1 + 3 x ) = +∞ and limx→0+ 1 x = +∞. Hence by l’Hôpital’s rule, lim x→0+ x ln(1 + 3 x ) = lim x→0+ (1 + 3 x )−1 (−3 x2 ) − 1 x2 = lim x→0+ 3(1 + 3 x )−1 = lim x→0+ 3x x+ 3 = 0.  Indeterminate Forms Involving Subtraction There are two basic cases here, we’ll do an example of each. Example 1.8 (∞–∞). Compute lim x→0 (cot(x)− csc(x)) .  6 Answer. Here we simply need to write each term as a fraction, lim x→0 (cot(x)− csc(x)) = lim x→0 ( cos(x) sin(x) − 1 sin(x) ) = lim x→0 cos(x)− 1 sin(x) Setting f(x) = cos(x) − 1 and g(x) = sin(x), both functions are differentiable near zero and lim x→0 (cos(x)− 1) = lim x→0 sin(x) = 0. We may now apply l’Hôpital’s rule. Write f ′(x) = − sin(x) and g′(x) = cos(x), so lim x→0 (cot(x)− csc(x)) = lim x→0 cos(x)− 1 sin(x) = lim x→0 − sin(x) cos(x) = 0.  Sometimes one must be slightly more clever. Example 1.9 (∞–∞). Compute lim x→∞ (√ x2 + x− x ) .  Answer. Again, this doesn’t appear to be suitable for l’Hôpital’s Rule. A bit of algebraic manipulation will help. Write lim x→∞ (√ x2 + x− x ) = lim x→∞ ( x (√ 1 + 1/x− 1 )) = lim x→∞ √ 1 + 1/x− 1 x−1 Now set f(x) = √ 1 + 1/x − 1, g(x) = x−1. Since both functions are differentiable for large values of x and lim x→∞ ( √ 1 + 1/x− 1) = lim x→∞ x−1 = 0, we may apply l’Hôpital’s rule. Write f ′(x) = (1/2)(1 + 1/x)−1/2 · (−x−2) and g′(x) = −x−2 7 So now look at the limit of the exponent lim x→0+ (lnx)(sinx) = lim x→0+ lnx 1 sinx . Let f(x) = lnx, g(x) = 1 sinx . Apply l’Hôpital’s rule, the limit is lim x→0+ f ′(x) g′(x) = lim x→0+ 1/x cosx sin2 x = lim x→0+ − sin2 x x cosx Apply l’Hôpital’s rule again, the above is = lim x→0+ − 2 sinx cosx cosx− x sinx = 0. Hence lim x→0+ xsinx = e0 = 1.  Example 1.12 (∞0). lim x→+∞ (ex + x) 1 x .  Answer. lim x→+∞ (ex + x) 1 x = lim x→+∞ e ln(ex+x) x . So now look at the limit of the exponent and apply l’Hôpital’s rule, the limit is lim x→+∞ ln(ex + x) x = lim x→+∞ ex ex + 1 . Apply l’Hôpital’s rule again, the limit is = lim x→+∞ ex ex = 1. Hence lim x→+∞ (ex + x) 1 x = e1 = e.  10 Exercise 1.1. limx→0 cosx−1 sinx answer. 0 Exercise 1.2. limx→∞ ex x3 answer. ∞ Exercise 1.3. limx→∞ √ x2 + x− √ x2 − x answer. 1 Exercise 1.4. limx→∞ lnx x answer. 0 Exercise 1.5. limx→∞ lnx√ x answer. 0 Exercise 1.6. limx→∞ ex+e−x ex−e−x answer. 1 Exercise 1.7. limx→0 √ 9+x−3 x answer. 1/6 Exercise 1.8. limt→1+ (1/t)−1 t2−2t+1 answer. −∞ Exercise 1.9. limx→2 2− √ x+2 4−x2 answer. 1/16 Exercise 1.10. limt→∞ t+5−2/t−1/t3 3t+12−1/t2 answer. 1/3 Exercise 1.11. limy→∞ √ y+1+ √ y−1 y answer. 0 Exercise 1.12. limx→1 √ x−1 3√x−1 answer. 3/2 Exercise 1.13. limx→0 (1−x)1/4−1 x answer. −1/4 Exercise 1.14. limt→0 ( t+ 1 t ) ((4− t)3/2 − 8) answer. −3 Exercise 1.15. limt→0+ ( 1 t + 1√ t ) ( √ t+ 1− 1) answer. 1/2 Exercise 1.16. limx→0 x2√ 2x+1−1 answer. 0 Exercise 1.17. limu→1 (u−1)3 (1/u)−u2+3/u−3 answer. 0 Exercise 1.18. limx→0 2+(1/x) 3−(2/x) answer. −1/2 Exercise 1.19. limx→0+ 1+5/ √ x 2+1/ √ x answer. 5 Exercise 1.20. limx→0+ 3+x−1/2+x−1 2+4x−1/2 answer. ∞ Exercise 1.21. limx→∞ x+x1/2+x1/3 x2/3+x1/4 answer. ∞ Exercise 1.22. limt→∞ 1− √ t t+1 2− √ 4t+1 t+2 answer. 2/7 Exercise 1.23. limt→∞ 1− t t−1 1− √ t t−1 answer. 2 Exercise 1.24. limx→−∞ x+x−1 1+ √ 1−x answer. −∞ 11