Indeterminate Forms - L'Hospital's Rule

Lecture 34 18.01 Fall 2006 Lecture 34: Indeterminate Forms - L’Hôpital’s Rule L’Hôpital’s Rule (Two correct spellings: “L’Hôpital” and “L’Hospital”) Sometimes, we run into indeterminate forms. These are things like 0 0 and ∞ ∞ For instance, how do you deal with the following? x3 − 1 0 = ?? 2 x→ 1 x − 1 0 lim Example 0. One way of dealing with this is to use algebra to simplify things: x3 − 1 (x − 1)(x2 + x + 1) x2 + x + 1 3 = lim = lim = 2 x→1 x − 1 x→1 x→1 (x − 1)(x + 1) x+1 2 lim In general, when f (a) = g(a) = 0, f (x) f (x) − f (a) lim f (x) f � (a) x→a x−a lim = lim x − a = = � x→a g(x) x→a g(x) g(x) − g(a) g (a) lim x→a x−a x−a This is the easy version of L’Hôpital’s rule: f (x) f � (a) = � x→a g(x) g (a) lim Note: this only works when g � (a) = � 0! In example 0, f (x) = x3 = 1; g(x) = x2 − 1 f � (x) = 3x2 ; g � (x) = 2x =⇒ f � (1) = 3; g � (1) = 2 The limit is f � (1)/g � (1) = 3/2. Now, let’s go on to the full L’Hôpital rule. 1 Lecture 34 18.01 Fall 2006 Example 1. Apply L’Hôpital’s rule (a.k.a. “L’Hop”) to x15 − 1 x→ 1 x3 − 1 lim to get x15 − 1 15x14 15 = lim = =5 3 2 x→1 x − 1 x→1 3x 3 Let’s compare this with the answer we’d get if we used linear approximation techniques, instead of L’Hôpital’s rule: x15 − 1 ≈ 15(x − 1) lim (Here, f (x) = x15 − 1, a = 1, f (a) = b = 0, m = f � (1) = 15, and f (x) ≈ m(x − a) + b.) Similarly, x3 − 1 ≈ 3(x − 1) Therefore, x15 − 1 15(x − 1) ≈ =5 x3 − 1 3(x − 1) Example 2. Apply L’Hop to lim x→ 0 sin 3x x to get lim x→ 0 This is the same as 3 cos 3x =3 1 � � d � � sin(3x) � = 3 cos(3x) � =3 dx x=0 x=0 Example 3. limπ x→ 4 √ sin x − cos x cos x + sin x 1 1 = limπ =√ +√ = 2 π x→ 4 x− 4 1 2 2 f (x) = sin x − cos x, f � (x) = cos x + sin x �π� √ f� = 2 4 Remark: Derivatives lim Δx→0 Δy 0 are always a type of limit. Δx 0 cos x − 1 . x Use L’Hôpital’s rule to evaluate the limit: Example 4. lim x→ 0 lim x→ 0 cos x − 1 − sin x = lim =0 x→0 x x 2 Lecture 34 18.01 Fall 2006 cos x − 1 . x→ 0 x2 cos x − 1 cos x − 1 − sin x − cos x 1 lim = lim = lim = lim =− x→0 x→0 x→0 x→0 x2 x2 2x 2 2 Just to check, let’s compare that answer to the one we would get if we used quadratic approximation techniques. Remember that: 1 cos x ≈ 1 − x2 (x ≈ 0) 2 1 2 1 1− x −1 (− )x2 cos x − 1 1 2 2 ≈ = =− x2 x2 2 x2 Example 5. lim Example 6. lim x→ 0 sin x . x2 sin x cos x = lim By L’Hôpital’s rule x→0 2x x2 If we apply L’Hôpital again, we get sin x lim − =0 x→0 2 But this doesn’t agree with what we get from taking the linear approximation: lim x→0 sin x x 1 ≈ 2 = → ∞ as x → 0+ x2 x x We can clear up this seeming paradox by noting that lim x→0 cos x 1 = 2x 0 0 The limit is not of the form , which means L’Hôpital’s rule cannot be used. The point is: look 0 before you L’Hôp! More “interesting” cases that work. ∞ , or if x → ∞, or x → −∞. Let’s ∞ apply this to rates of growth. Which function goes to ∞ faster: x, eax , or ln x? It is also okay to use L’Hôpital’s rule on limits of the form Example 7. For a > 0, eax aeax = lim = +∞ x→∞ x x→∞ 1 grows faster than x (for a > 0). lim So eax Example 8. eax aeax c2 eax a10 eax = by L’Hôpital = lim = lim = · · · = lim =∞ 10 9 8 x→∞ x x→∞ 10x x→∞ 10 · 9x x→∞ 10! lim 3 Lecture 34 18.01 Fall 2006 You can apply L’Hôpital’s rule ten times. There’s a better way, though: � eax x10 �1/10 = eax = lim x→∞ x10 x→∞ � lim eax/10 x eax/10 x �10 = ∞10 = ∞ Example 9. lim x→∞ ln x 1/x = lim = lim 3x−1/3 = 0 x→∞ 1/3x−2/3 x→∞ x1/3 Combining the preceding examples, ln x � x1/3 � x � x10 � eax (x → ∞, a > 0) 0 ∞ and . But, we sometimes face other indeterminate limits, such 0 ∞ ∞ 0 as 1 , 0 , and 0 · ∞. Use algebra, exponentials, and logarithms to put these in L’Hôpital form. L’Hôpital’s rule applies to Example 10. lim xx for x > 0. x→ 0 Because the exponent is a variable, use base e: lim xx = lim ex ln x x→ 0 x→ 0 First, we need to evaluate the limit of the exponent lim x ln x x→0 This limit has the form 0 · ∞. We want to put it in the form Let’s try to put it into the 0 form: 0 0 ∞ or . 0 ∞ x 1/ ln x 1 , though, so that approach isn’t helpful. ln x ∞ Instead, let’s try to put it into the form: ∞ We don’t know how to find lim x→0 ln x 1/x Using L’Hôpital’s rule, we find ln x 1/x = lim = lim (−x) = 0 x→0 1/x x→0 −1/x2 x→0 lim x ln x = lim x→ 0 Therefore, lim (x ln x) lim xx = lim ex ln x = e x→0 = e0 = 1 x→ 0 x→ 0 4