8.1 L’Hôpital’s Rule Actually, L’Hôpital’s Rule was developed by his teacher Johann Bernoulli. De L’Hôpital paid Bernoulli for private lessons, and then published the first Calculus book based on those lessons. Guillaume De l'Hôpital 1661 - 1704 8.1 L’Hôpital’s Rule Consider: x2 4 lim x2 x 2 If we try to evaluate this by direct substitution, we get: Zero divided by zero can not be evaluated, and is an example of indeterminate form. In this case, we can evaluate this limit by factoring and canceling: x2 4 lim x2 x 2 x 2 x 2 lim x 2 x2 lim x 2 4 x 2 0 0 8.1 L’Hôpital’s Rule f x lim xa g x x2 4 lim x 2 x 2 x2 4 The limit is the ratio of the numerator over the denominator as x approaches 2. 4 0.05 3 2 1 -3 -2 -1 1 0 -1 -2 x 2 3 If we zoom in far enough, the curves will2 appear as2.05 0 1.95 x straight lines. -3 x2 -4 -0.05 -5 8.1 L’Hôpital’s Rule 0.05 f x lim xa g x x2 4 lim x 2 x 2 As f x g x df x 2 becomes: dg 0 1.95 -0.05 2 dx x 2.05 df df dx dg dg dx 8.1 L’Hôpital’s Rule f x lim xa g x d 2 x 4 x2 4 dx lim lim x 2 x 2 x 2 d x 2 dx 2x lim 4 x2 1 8.1 L’Hôpital’s Rule L’Hôpital’s Rule: f x If lim is indeterminate, then: xa g x f x lim lim x a g x x a f x g x 8.1 L’Hôpital’s Rule We can confirm L’Hôpital’s rule by working backwards, and using the definition of derivative: f a g a f x f a lim xa xa g x g a lim xa xa f x f a lim x a g x g a f x f a xa lim xa g x g a xa f x 0 f x lim lim xa g x 0 x a g x 8.1 L’Hôpital’s Rule Example: 1 cos x lim sin x 0 lim x 0 2 1 2 x x 0 x x If it’s no longer indeterminate, then STOP! If we try to continue with L’Hôpital’s rule: sin x x 0 1 2 x lim cos x 1 x 0 2 2 lim which is wrong, wrong, wrong! 8.1 L’Hôpital’s Rule On the other hand, you can apply L’Hôpital’s rule as many times as necessary as long as the fraction is still indeterminate: x 1 x 1 2 lim x 0 x2 1 2 1 1 x 1 x 2 lim x 0 x2 0 0 1 1 1 2 1 x 2 lim 2 x 0 2x 3 1 1 x 2 not lim 4 x 0 2 1 4 2 1 8 0 0 0 0 8.1 L’Hôpital’s Rule L’Hôpital’s rule can be used to evaluate other indeterminate forms besides 0 . 0 The following are also considered indeterminate: 0 1 0 The first one, , can be evaluated just like The others must be changed to fractions first. 0 0 . 0 0 8.1 L’Hôpital’s Rule 1 This approaches lim x sin 0 x x 1 sin x 0 lim This approaches x 1 0 x 1 1 1 cos 2 sin x x lim cos 1 x cos 0 1 lim lim x x x 1 1 x 2 x x 8.1 L’Hôpital’s Rule Indeterminate Forms: lim f x L x a 1/ x lim x x lim ln x1 / x e x 0 e e 1 0 00 lim f x lim e x a x a 1 lim ln x x x e ln x lim x x 1 lim x x 1 e 0 ln f x eL L’Hôpital applied 1 8.2 Relative Rates of Growth x y e The function grows very fast. If x is 3 inches, y is about 20 inches: 35 30 25 20 15 10 5 3, 20 0 20 We have gonethe lessy-value than halfAt 64 inches, way across the board would be at the edge of the horizontally, and already the yknown universe! value the (10.5 would billion reach light-years) Andromeda Galaxy! 40 60 At x 10 inches, At x 44 inches, 1 y mile y 2 million light-years 3 80 8.2 Relative Rates of Growth The function y = ln x grows very slowly. We would have to move 2.6 miles to the right before the line moves a foot above the x-axis! 35 30 25 20 15 10 5 0 By the time we reach the edge of the universe again (10.5 billion light-years) the chalk line will only have reached 64 inches! 20 40 60 80 The function y = ln x increases everywhere, even though it increases extremely slowly. 8.2 Relative Rates of Growth Let f (x) and g(x) be positive for x sufficiently large. Definitions: Faster, Slower, Same-rate Growth as x 1. f grows faster than g (and g grows slower than f ) as x if lim x f x g x or lim x g x f x 2. f and g grow at the same rate as lim x f x g x L0 0 if x 8.2 Relative Rates of Growth WARNING Please temporarily suspend your common sense. 8.2 Relative Rates of Growth According to this definition, y = 2x does not grow faster than yx 2x lim lim 2 2 6 x x x 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 yx y 2x Since this is a finite non-zero limit, the functions grow at the same rate! 1 2 3 4 5 6 The book says that “f grows faster than g” means that for large x values, g is negligible compared to f. 8.2 Relative Rates of Growth Which grows faster, ex lim 2 x x x 2 eor x? Weiscan confirm thisso graphically: This indeterminate, we 10 apply L’Hôpital’s rule. 9 e lim x 2 x x ex y 2 x 8 7 Still indeterminate. 6 5 ex lim x 2 4 3 2 1 e x grows faster than 2 x. 0 1 2 3 4 5 6 7 8 9 10 8.2 Relative Rates of Growth “Growing at the same rate” is transitive. In other words, if two functions grow at the same rate as a third function, then the first two functions grow at the same rate. 8.2 Relative Rates of Growth Show that f x x 5 and g x 2 x 1 grow 2 at the same rate as x . 2 8.2 Relative Rates of Growth Let h x x x 5 lim lim x x x 2 lim x 2 x 1 x 2 x2 5 lim x x2 x2 5 lim 2 2 x x x 2 x 1 x 2 2 f f h 1 1 lim lim 1 x g x h g 4 4 50 lim 1 2 1 x x 2 2 x 1 0 lim 4 x x x f and g grow at the same rate. 8.2 Relative Rates of Growth Definition f of Smaller Order than g Let f and g be positive for x sufficiently large. Then f is of smaller order than g as x if lim x We write f x g x 0 f o g and say “f is little-oh of g.” Saying f o g is another way to say that f grows slower than g. 8.2 Relative Rates of Growth Definition f of at Most the Order of g Let f and g be positive for x sufficiently large. Then f is of at most the order of g as x if there is a positive integer M for which f x g x M for x sufficiently large f O g and say “f is big-oh of g.” Saying f O g is another way to say that f grows no faster than g. We write 8.3 Improper Integrals Until now we have been finding integrals of continuous functions over closed intervals. Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals. 8.3 Improper Integrals 4 1 0 1 x dx 1 x The function is undefined at x = 1 . Since x = 1 is an asymptote, the function has no maximum. We could define this integral as: lim b 1 b 0 1 x dx 1 x 3 Can we find the area under an infinitely high curve? 2 1 0 1 (left hand limit) We must approach the limit from inside the interval. 8.3 Improper Integrals lim b 1 b 0 1 x dx 1 x 1 x 1 x dx 1 x 1 x 1+x 1 x2 dx 1 1 sin 1 x u 2 du 2 1 1 x2 dx x 1 x2 u 1 x2 du 2x dx 1 du x dx 2 dx 8.3 Improper Integrals 1 1 x2 x dx 1 x2 1 2 1 sin x u du 21 1 sin 1 x u 2 b dx u 1 x2 du 2x dx 1 du x dx 2 This integral converges because it approaches a solution. lim sin 1 x 1 x 2 b 1 0 0 0 2 1 lim sin b 1 b 2 sin 1 0 1 b 1 2 1 8.3 Improper Integrals 4 1 lim dx b 0 b x dx 0 x 1 1 3 2 1 lim ln x b b0 lim ln1 ln b 1 b 0 -1 1 lim ln b 0 b 0 1 -1 This integral diverges. 8.3 Improper Integrals 4 dx 3 0 x 1 x 1 3 The function approaches when x 1 . 2 3 2 3 0 b 1 b1 2 3 dx lim x 1 3 c 1 0 lim 3 x 1 2 dx lim x 1 b 3 1 b 3 0 c lim 3 x 1 c 1 1 3 3 c 2 3 1 dx 0 1 2 3 8.3 Improper Integrals lim x 1 b b 1 2 3 dx lim x 1 3 c 1 0 lim 3 x 1 1 b 3 b1 c lim 3 x 1 0 0 c 1 2 3 dx 1 3 3 c 0 lim 3 b 1 3 1 lim 3 2 3 c 1 b 1 c1 1 3 1 3 1 3 1 3 3 33 2 8.3 Improper Integrals 1 dx P x b P 1 1 P 1 lim b P 1 P 1 P0 (P is a constant.) 1 x P dx b lim x b 1 P What happens here? P 1 b If P 1 then gets bigger and bigger as b , therefore the integral diverges. dx b 1 P 1 lim x b P 1 1 If P 1 then b has a negative exponent and b P 1 0 , therefore the integral converges. 8.3 Improper Integrals 1 x e dx b Converges lim e dx x lim e b 1 lim e e b b x b b 1 lim b 1 0 1 1 e b e 1 e 8.3 Improper Integrals Does 1 e x2 dx converge? Compare: 1 e x 2 to 1 ex For x 1, e for positive values of x. x2 e x 1 e x2 1 x e 8.3 Improper Integrals For x 1, e x2 e x 1 e Since 1 e x2 x2 1 x e 1 is always below , x e we say that it is “bounded above” by 1 x e 1 1 Since x converges to a finite number, x 2 must also converge! e e 8.3 Improper Integrals Direct Comparison Test: Let f and g be continuous on a, with 0 f x g x for all xa , then: 1 f x dx a 2 g x dx a g x dx converges. converges if diverges if a f x dx diverges. a 8.3 Improper Integrals 1 sin 2 x dx 2 x The maximum value of sin x 1 so: 2 sin x 1 0 2 2 x x on on 1, 1 sin 2 x Since 2 converges, converges. 2 x x 8.3 Improper Integrals 1 1 x 2 0.1 x2 0.1 x 1 dx for positive values of x, so: 1 on 2 x 0.1 x 1 Since diverges, x 1, 1 x 0.1 2 diverges. 8.3 Improper Integrals Does 1 If functions grow at the same rate, then either they both converge or both diverge. dx 2 converge? 1 x x the “1” in the denominator becomes 1 insignificant, so we compare to 2 . x 1 1 Since 2 converges, 2 2 1 x x x 2x lim lim 1 As x 1 1 x2 x x2 lim x 2x 1 1 x2 converges. 8.3 Improper Integrals 1 dx 1 x2 lim b b 1 1 b dx 1 x2 2 4 1 lim tan x lim tan b tan 1 b 1 2 1 b 4 2 4 4 As y , x 2 y tan x 8.3 Improper Integrals 1 dx 1 x2 lim b b 1 1 dx 1 x2 1 lim tan x b lim x 2 dx b 1 b 1 2 4 1 dx 2 x b lim tan 1 b tan 1 1 b 4 lim x 1 b b 1 0 1 1 lim b b 1 1 8.4 Partial Fractions 5x 3 x 2 2 x 3 dx 5x 3 x 3 x 1 This would be a lot easier if we could re-write it as two separate terms. A B x 3 x 1 Multiply by the common denominator. 5x 3 A x 1 B x 3 5x 3 Ax A Bx B 3 5x Ax Bx 5 A B 3 A B 3 3 A 3B Set like-terms equal to each other. Solve two equations with two unknowns. 8.4 Partial Fractions 5x 3 x 2 2 x 3 dx A B 5x 3 x 3 x 1 x 3 x 1 5x 3 A x 1 B x 3 5x 3 Ax A Bx B 3 5x Ax Bx 5 A B 3 A B 3 3 A 3B 5 A B 3 A 3B 3 A 3B 8 4B 5 A 2 2B 3 A 3 2 x 3 x 1 dx 3ln x 3 2ln x 1 C Solve two equations with Thisunknowns. technique is called two Partial Fractions 8.4 Partial Fractions Good News! The AP Exam only requires non-repeating linear factors! The more complicated methods of partial fractions are good to know, and you might see them in college, but they will not be on the AP exam or on my exam. 8.4 Partial Fractions 6x 7 x 2 2 A B 2 x 2 x 2 6 x 7 A x 2 B Repeated roots: we must use two terms for partial fractions. 7 12 B 6x 7 Ax 2 A B 6x Ax 7 2A B 6 A 7 26 B 5 B 6 5 x 2 x 2 2 8.4 Partial Fractions 2 x3 4 x 2 x 3 x2 2 x 3 If the degree of the numerator is higher than the degree of the denominator, use long division first. 2x x 2 2 x 3 2 x3 4 x 2 x 3 2 x3 4 x 2 6 x 5x 3 5x 3 2x 2 x 2x 3 5x 3 2x x 3 x 1 3 2 2x x 3 x 1 (from example one) 8.4 Partial Fractions x 2 x 4 2 1 x 1 irreducible quadratic factor 2 Ax B C D 2 x 1 x 1 x 12 repeated root 2 x 4 Ax B x 1 C x 2 1 x 1 D x 2 1 2 2 x 4 Ax B x 2 2 x 1 C x 3 x 2 x 1 Dx 2 D 2 x 4 Ax3 2 Ax 2 Ax Bx 2 2 Bx B Cx 3 Cx 2 Cx C Dx 2 D 8.4 Partial Fractions dx 4 x2 These are in the same form. 22 2 4 x a x 2sec d 2sec 2 sec d ln sec tan C 4 x x C 2 2 2 ln x 2a 4 x2 sec 2 2sec 4 x 2 x tan 2 2 tan x 2sec 2 d dx 8.4 Partial Fractions dx 4 x2 2sec2 d 2sec 4 x2 x C 2 ln ln 4 x 2 x ln 2 C sec d ln sec tan C ln 4 x2 x C 2 2 This is a constant. ln 4 x2 x C 8.4 Partial Fractions a2 x2 a2 x2, If the integral contains we use the triangle at right. If we need a2 x,2 we move a to the hypotenuse. a a2 x2 x a If we need x2 a,2 we move x to the hypotenuse. x x x a 2 a 2 8.4 Partial Fractions x 2 dx 3 9 x2 9 x2 9sin 2 3cos d 3cos 1 cos 2 9 d 2 9 1 cos 2 d 2 9 9 1 sin 2 C 2 2 2 x x sin 3 3sin x 9 x2 cos 3 3cos 9 x2 3cos d dx x 1 x sin sin 3 3 9 1 x 9 x 9 x 2 sin C 2 3 2 3 3 8.4 Partial Fractions 3 x 2 dx 9 x 2 x 9 x2 9 1 x x sin 9 x2 C 2 3 2 9 9 1 sin 2 C 2 2 2 9 1 x 9 sin 2sin cos C 2 3 4 9 1 x 9 x 9 x 2 sin C 2 3 2 3 3 8.4 Partial Fractions dx We can get 2x x 2 into the necessary form by completing the square. 2x x2 2x x 2 x 2x 2 x2 2x 1 1 x 1 1 2 1 x 1 2 sin u dx 1 x 1 2 cos d du 8.4 Partial Fractions dx 2x x du 1 u2 cos d cos d 2 dx 1 x 1 2 1 Let u x 1 du dx 1 sin u C C sin 1 x 1 C u 1 u2 1 u2 1 u2 cos 1 sin u cos d du 8.4 Partial Fractions dx 4 x2 4 x 2 Complete the square: dx 4 x2 4 x 1 1 2 x 1 2 1 4 x2 4 x 2 2 x 1 2 1 8.4 Partial Fractions dx 4 x2 4 x 2 dx 2 x 1 2 1 1 du 2 u2 1 1 sec 2 d 2 sec 2 Let u 2x 1 du 2 dx 1 du dx 2 u2 1 u 1 tan u sec 2 d du sec u 2 1 1 1 2 2 d C sec u 1 2 2 1 1 1 tan 1 2 x 1 C tan u C 2 2 8.4 Partial Fractions Here are a couple of shortcuts that are result from Trigonometric Substitution: du 1 1 u u 2 a 2 a tan a C du u sin C a a2 u 2 1 These are on your list of formulas. They are not really new.