Math òýA. Lecture Ôý.

′ ′ ′ eorem Ô (L’Hopital’s Rule) Suppose that f (x) and g (x) exist with g (x)≠ ý for x ≠ xý in an open interval containing xý, that either f (x) and g(x) both tend to ý or both tend to ±∞ as x → xý, ′ ′ and that f (x)~g (x) → L as x → xý. en

′ f (x) = f (x) = lim lim ′ L. x→xý g(x) x→xý g (x)

In this theorem, L may be a number or ±∞ and the condition x → xý may be replaced by x → ∞ or x → −∞ with appropriate changes in the intervals used in the denition. (is result is called L’Hopital’s rule because it was rst published in Analyse des Inniments Petits pour l’Intelligence des Lignes Courbes by the Marquis G. F. A. de L’Hopital in ÔâÀâ. is book was one of the rst textbooks on dišerential calculus. e result, however, was not proved by de l’Hopital but by another mathematician, Jean Bernoulli (brother of Jacob, father of Daniel), who received a regular salary from the marquis in exchange for the rights to Bernoulli’s mathematical discoveries.) f (x) ý e quotient in the theorem is called Z• †•ou±u§“†•Z±u €™§“ ™€ ±í£u if f (x) and g(x) g(x) ý ∞ tend to zero and an †•ou±u§“†•Z±u €™§“ ™€ ±í£u if f (x) and g(x) tend to ±∞. ∞ sin(òπx) Example Ô Use l’Hopital’s Rule to nd the limit lim . x→Ô xò − Ô sin(òπx) Answer: y = sin(òπx) → ý and y = xò − Ô → ý as x → Ô. ● e quotient is an indeterminate form of xò − Ô d d sin òπx cos òπx òπx ý sin(òπx) dx dx òπ cos òπx òπ type as x → Ô. ● lim = lim [ ( )] = lim ( ) ( ) = lim ( ) = = π ý x→Ô xò − Ô x→Ô d x→Ô òx x→Ô òx ò xò − Ô dx ( ) L’Hopital’s Rule has a special geometric interpretation in cases where the functions f and g have ′ contunuous derivatives at the limiting value xý of x and g (xý) is not zero. en the tangent lines ′ y = f (xý)+ f (xý)(x − xý) and y = g(xý)(x − xý) approximate the graphs closely near xý and f (xý) and g(xý) are zero, so that. for x near xý, ′ f (x) ≈ f (xý)(x − xý) ′ g(x) ≈ g (xý)(x − xý). ese approximations are accurate enough so that ′ ′ f (x) = f (xý)(x − xý) = f (xý) lim lim ′ ′ . x→xý g(x) x→xý g (xý)(x − xý) g (xý) us, the limit of the quotient of the functions as x → xý is equal to the ratio of the slopes of their tangent lines at xý. is is illustrated in Figures Ô through ç below for the functions of Example Ô. e tangent line to y = sin(òπx) at x = Ô in Figure Ô has slope òπ, the tangent line to y = xò − Ô at x = Ô in Figure ò has slope ò, sin(òπx) òπ and the limit as x → Ô of the quotient in Figure ç as x → ý is = π. xò − Ô ò Math òýA Lecture Ôý (Þ/ÔÀ/ÔÔ), p. ò

sin(òπx) y y = sin(òπx) y y = xò − Ô y y = xò − Ô ç ç ç

ò ò ò

Ô Ô

Ô òý x Ô ò x Ô x

òπ Tangent line of slope òπ Tangent line of slope ò y → = π as x → Ô ò FIGURE Ô FIGURE ò FIGURE ç

In the next example L’Hopital’s Rule is applied twice. xò − òx + Ô Example ò Find lim . x→Ô x − Ô − ln x xò − òx + Ô ý Answer: y = is an indeterminate form of type as x → Ô since y = xò − òx + Ô and y = x − Ô − ln x x − Ô − ln x ý tend to zero as x → Ô. ● d ò xò − òx + Ô x − òx + Ô òx − ò lim = lim dx ( ) = lim provided the limit on the right exists. ● x→Ô x − Ô − ln x x→Ô d x→Ô Ô − x−Ô x − Ô − ln x dx ( ) òx − ò ý is also indeterminant of type as x → Ô because òx − ò → ý and Ô − x−Ô → ý as x → Ô. ● Ô − x−Ô ý d xò − òx + Ô òx − ò òx − ò ò ò lim = lim = lim dx ( ) = lim = = ò x→Ô x − Ô − ln x x→Ô Ô − x−Ô x→Ô d x→Ô x−ò Ô−ò Ô − x−Ô dx ( ) Example ò is illustrated in Figures ¥ through â, which show the graphs of the functions y = xò − òx + Ô and y = x − Ô − ln x and their quotient. Notice that in this case we cannot express the limit of the quotient of the functions as the quotient of the slopes of the tangent lines to their graphs because the tangent lines (the x-axis) are horizontal and their slopes are zero.

y y y

ç

ò ò

Ô ò x Ô x Ô x

xò − òx + Ô y = xò − òx + Ô y = x − Ô − ln x y = x − Ô − ln x FIGURE ¥ FIGURE FIGURE â Lecture Ôý (Þ/ÔÀ/ÔÔ), p. ç Math òýA

In the next example, L’Hopital’s Rule does not apply and cannot be used. sin(çx) Example ç What is lim ? x→ý Ô + sin(¥x) Answer: e numerator y = sin çx tends to ý as x → ý, but the denominator y = Ô + sin ¥x tends to the nonzero ( ) ( ) lim sin çx sin çx x→ý ý number Ô. ● e quotient is not indeterminate. ● lim ( ) = ( ) = = ý x→ý Ô + sin ¥x lim Ô + sin ¥x Ô ( ) x→ý [ ( )] Mistakenly applying l’Hopital’s Rule to Example ç would lead to the incorrect result,

d d sin çx cos çx çx sin(çx) dx [ ( )] ( )dx ( ) ç cos çx lim = lim = lim = lim ( ) = ç . x→ý Ô + sin(¥x) x→ý d x→ý d x→ý ¥ cos ¥x ¥ Ô + sin ¥x sin ¥x ¥x ( ) dx [ ( )] ( )dx ) ex~ç Example ¥ Find lim . x→∞ x ex~ç ∞ Answer: y = is an indeterminate of type as x → ∞ because ex~ò and x both tend to ∞. ● x ∞ d x~ò x~ò d Ô x~ò e e x e ò lim = lim dx ( ) = lim dx ( ) = lim Ô ex~ò =∞ x→∞ x x→∞ d x→∞ Ô x→∞ ò x dx ( ) is example ilustrates the general principle that exponential functions y = eax for a > ý tend to ∞ faster than all polynomials as x → ∞. ex~ò Example Sketch the graph of y = by studying the function and its rst derivative. x ex~ò Answer: Study the function. ● y = is dened and continuous for x ≠ ý, is positive for x > ý and is negative x for x < ý. ● It tends to ∞ as x → ∞ by Example ¥. ● It tends to ý as x → −∞ because ex and Ô x both tend to ~ zero. ● It tends to −∞ as x → ý− because ex~ò → Ô and Ô x → −∞. ● It tends to ∞ as x → ý+ because ex~ò → Ô ~ and Ô x → ∞. ● Figure Þ ● ~

ý y y → −∞ ∞ y y → ∞ ← ← x FIGURE Þ y < ý ý y > ý

Study the derivative:

d x~ò x~ò d x~ò d Ô x~ò x~ò x e − e x xe x − e Ô d e ò x − Ô x − ò y′ = = dx ( ) dx ( ) = dx ‰ Ž = ò eò~x = eò~x ● dx Œ x ‘ xò xò Œ xò ‘ ‹ òxò  e derivative is not dened at x = ý, is zero at x = ò, is negative for x < ý and for ý < x < ò, and is positive for x > ý. ● e function is decreasing on −∞, ý and on ý, ò , is increasing on ò, ∞ , and has a local minimum at x = ò. ● Figure — ( ) ( ] [ )

y ↳ y ↳ y ↱

ý ò x FIGURE — y′ < ý y′ < ý y′ > ý Math òýA Lecture Ôý (Þ/ÔÀ/ÔÔ), p. ¥

Draw the graph. ● One approach: Use a calculator to nd the values y −Ô = −e−Ô~ò ≐ −ý.âÔ, y Ô = eÔ~ò ≐ Ô.â , ( ) ( ) y ò = Ô e ≐ Ô.çâ, and y â = Ô eç ≐ ç.ç and plot these points. ● Figure À ( ) ò ( ) â y ex~ò y = — x

¥

ò ¥ â x FIGURE À

ln x Example â Show that → ý as x → ∞ for every positive integer n. √n x ln x Answer: y = is an indeterminate form of type ∞ ∞ as x → ∞ because ln x → ∞ and √n x → ∞ as x → ∞. ● √n x ~ d ln x ln x x−Ô x−Ô n lim = lim dx ( ) = lim = lim = lim = ý x→∞ xÔ~n x→∞ d x→∞ Ô x→∞ Ô x→∞ xÔ~n xÔ~n x(Ô~n)−Ô x(Ô~n) x−Ô dx ( ) n n Indeterminate forms of types ý ⋅ ∞ e product y = f x g x of a function f that tends to ý and a function g that tends to ∞ is called an †•ou±u§“†•Z±u €™§“( ) ™€( ) ±í£u ý ⋅ ∞. e limits of many such expressions can be found by rewriting them as indeterminate forms of type ý ý or ∞ ∞. ~ ~ Example Þ Find lim √x ln x . x→ý+ [ ( )] Answer: y = √x ln x is indeterminate of type ý ⋅∞ as x → ý+ because y = √x → ý and y = ln x → −∞. ● ( ln) x Rewrite √x ln x as , which is an indeterminate form of type ∞ ∞ as x → ý+. ● ( ) x−Ô~ò ~ d ln x ln x x−Ô lim √x ln x = lim = lim dx ( ) = lim = lim −òxÔ~ò = ý x→ý+ [ ( )] x→ý+ x−Ô~ò x→ý+ d x→ý+ − Ô x−ç~ò x→ý+ ( ) x−Ô~ò ò dx ( ) √x We could attempt to nd the limit in Example Þ by writing the given expression in the form , ln x −Ô which is of type ý ý as x → ý+. en, however, it does not help to apply l’Hopital’s Rule because the( quotient) ~ d Ô~ò x Ô x−Ô~ò −√x of the derivatives is dx ( ) = ò = = − Ô √x ln x ò, and we cannot tell what its d − ln x −òx−Ô ò ln x −ò ò ( ) ln x −Ô ( ) ( ) dx [( ) )] limit is as x → ý+. Nor would it help to apply l’Hopital’s Rule to the new quotient. Lecture Ôý (Þ/ÔÀ/ÔÔ), p. Math òýA

Indeterminate forms of types ýý, Ô∞ and ∞ý e expression y = f x g(x) is †•ou±u§“†•Z±u ™€ ±í£u ýý if f x → ý and g x → ý; it is ™€ ±í£u Ô∞ if ( ) ( ) ( ) f x → Ô and g x → ±∞; and it is ™€ ±í£u ∞ý if f x → ∞ and g x → ý. e limits of such expressions can( ) oŸen by found( ) by applying l’Hopital’s Rule to their( ) logarithms. ( ) Example — Find the limit of y = x−x as x → ý+. Answer: y = x−x is of type ýý as x → ý+. ● ln x−x = −x ln x is an indeterminate of type ý ⋅∞. ● ( ) d − ln x ln x −x−Ô lim ln x−x = lim −x ln x = lim = lim dx ( ) = lim = lim x = ý ● Because its x→ý+ ( ) x→ý+ ( ) x→ý+ x−Ô x→ý+ d x→ý+ −x−ò x→ý+ x−Ô dx ( ) −x logarithm tends to ý, the function y = x−x = eln(x ) tends to eý = Ô as x → ý+/ ● lim x−x = Ô ● e curve x→ý+ y = x−x is shown in Figure Ôý.

y y = x−x

Ô

FIGURE Ôý Ô x

Piecewise constant rates of change Example À At Ôý:ýý AM one morning a truck driver is ý miles east of a town He drives Þ miles per hour toward the east for two hours to make a delivery. Next, he drives west at ý miles per hour for two hours to make another delivery and then drives east at ý miles per hour for two more hours. According to this mathematical model, his velocity toward the east is the “step” function of Figure ÔÔ with t = ý at Ôý AM. (a) How far is he from the rst town at t = â? (b) How is the answer to part (a) related to the areas of the rectangles in Figure Ôò?

v (miles per hour) v = v(t) v (miles per hour) Þ Þ

A ò ò C

Ôòç¥ â t Ô â t B (hours) (hours) − ý − ý

FIGURE ÔÔ FIGURE Ôò

Answer: (a) At t = â, he is [ ý miles] + [Þ miles per hour][ò hours] − [ ý miles per hour][ò hours] + [ ý miles per hour][ò hours] = ý + Ô ý − Ôýý + Ôýý = òýý miles east of the town. (b) His location at t = â equals his location at t = ý plus the the sum of the areas of rectangles A and C in Figure Ôò, minus the area of rectangle B. Math òýA Lecture Ôý (Þ/ÔÀ/ÔÔ), p. â

Partitions and step functions A £Z§±†±†™• of a nite closed interval a, b is a nite number of points xý, xÔ, xò, xç,..., xN such that [ ] a = xý < xÔ < xò < xç⋯ < xN = b. ese points divide a, b into N subintervals. Figure Ôç shows, for example, the four subintervals [ ] xý, xÔ , xÔ, xò , xò, xç , and xç, x¥ that are dened by a partition, a = xý < xÔ < xò < xç < x¥ = b. [ ] [ ] [ ] [ ]

a = xý xÔ xò xç x¥ = b x

FIGURE Ôç

A function is a «±u£ €¶•h±†™• on an interval a, b if it is constant on the interiors xý, xÔ , xÔ, xò , [ ] ( ) ( ) xò, xç ,..., xN−Ô, xN of the subintervals in a partition of the interval. e function might or might not be dened( ) at endpoints( of) the subintervals.

eorem ò Suppose that a function y = F x is continuous on a nite closed interval a, b and that its ( ) [ ] derivative r = F′ x is a step function on a, b . en the region between the graph r = F′ x and the x-axis for( a)≤ x ≤ b consists of a nite[ ] number of rectangles, and the change in the( ) function’s value from x = a to x = b is given by e area of e area of ⎡ ⎤ ⎡ ⎤ − = ⎢ all rectangles ⎥ − ⎢ all rectangles ⎥ F b F a ⎢ ⎥ ⎢ ⎥ . ( ) ( ) ⎢ above the x-axis ⎥ ⎢ below the x-axis⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ Example Ôý A water tank contains çýý gallons of water at time t = ý (minutes) and the rate of žow r = r t (gallons per minute) into the tank for ý ≤ t ≤ Þý is the step function in Figure Ô¥. How much( ) water is in the tank at t = Þý?

r (gallons per minute) r = r(t) r (gallons per minute) Ôý Ôý

Ôý òý çý ¥ý ý âý Þý t òý çý ý Þý t − (minutes) − (minutes) −Ôý −Ôý

FIGURE Ô¥ FIGURE Ô

Answer: We assume that the volume V t of water in the tank at time t is continuous on ý, Þý . ● gallons ( ) gallons gallons [ ] V Þý − V ý = Ôý Ôý minutes + Ôý minutes − çý minutes ( ) ( )  minute [ ]  minute [ ]  minute [ ] gallons − Ôý òý minutes = Ôýý + ý − Ô ý − òýý gallons = −òýý gallons. ●  minute [ ] V Þý = V ý − òýý = çýý − òýý = Ôýý ● ere are Ôýý gallons of water in the tank at t = Þý ( ) ( ) e volume of water in the tank at t = Þý in Example Ôý is equal to the volume at t = ý plus the area of the two rectangles above the t-axis in Figure Ô , minus the area of the two rectangles below the t-axis. Lecture Ôý (Þ/ÔÀ/ÔÔ), p. Þ Math òýA

Example ÔÔ What is F Ô if y = F x is continuous on ý,Ô , F ý = Ôý, and F′ x equals Ôý for ý < x < Ô , ( ) ( ) [ ] ( ) ( ) ç Ô < < ò Ô < < equals òý for ç x ç , and equals çý for ç x Ô? Answer: F Ô = F ý + Ôý Ô + òý Ô + çý Ô = Ôý + âý = çý ( ) ( ) ( ç ) ( ç ) ( ç ) ç Example Ôò A few plants, including jack in the pulpit, skunk cabbage, and sacred lotus, can create heat from oxygen and other nutrients. e step function in Figure Ôâis amodelof therate of consumption of oxygen by a sacred lotus plant from noon one day to midnight çâ hours later. e žower kept its temperature between çý○C and çÞ○C as the air temperature varied between Ôý○C and ç ○C by consuming more oxygen in the colder periods. (Data adapted from Scientic American) How much oxygen did the žower consume during the çâ hours? (t is measured in minutes.)

r (milliliters per minute) ç r = r(t) ò. ò.¥ Ô.À ò.ý ò Ô.Þ Ô.ý Ô òPM Ôò PM â AM Ôò AM â PM Ôò PM â AM Ôò t FIGURE Ôâ

Answer: [Total oxygen consumed]=Ô çâý + Ô.À çâý + ò. çâý + ò.¥ çâý + Ô.Þ çâý + ò.ý çâý = ¥Ô¥ý milliliters ( ) ( ) ( ) ( ) ( ) ( ) Example Ôç Figures ÔÞ and ԗ show the graphs of a piecewise linear function y = f x and its derivative ( ) r = f ′ x . According to the conclusion of eorem ò, f ç − f ý is equal to the combined area ò(Ô )+ Ô ò = ¥ of rectangles A and B in Figure ԗ. Instead,( ) (f ç) − f ý = ò − ý = ò. Explain. ( ) ( ) ( ) ( )

y y = f (x) r r = f ′(x) ò ò

Ô Ô B A

Ôò ç x Ôò ç x

FIGURE ÔÞ FIGURE ԗ

Answer: eorem ò does not apply because f is not continuous on ý.ç . [ ] Math òýA Lecture Ôý (Þ/ÔÀ/ÔÔ), p. —

Approximating continuous rates of change with step functions Suppose that the continuous function v = v t of Figure ÔÀ is the velocity of a car that is at s = s t on an ( ) ( ) s-axis at time t, so that v t = s′ t for a < t < b and that s = s t is continuous on a, b . Notice that the region between the graph( and) the( )t axis for a ≤ t ≤ b in Figure( ÔÀ) is in two parts. e[ region] labeled A for the time period a ≤ t < c, when the car’s velocity is positive, is above the t-axis. e region labeled B for c < t ≤ b, when the car’s velocity is negative, is below the t-axis.

v v v = v(t) v = v(t)

A b b a c t a t B

FIGURE ÔÀ FIGURE òý

We cannot apply eorem ò in this case because the velocity is not a step function. Instead we approximate v = v t with a step function by approximating regions A and B by rectangles, as in Figure òý, where the sides of( the) rectangles are determined by a partition of a, b and the tops are chosen to intersect the graph of v. If the car’s velocity were given by the step function,[ we could] apply eorem ò and conclude that the change in the car’s position s b − s a from t = a to t = b equals the area of the two rectangles above the t-axis, minus the area of the( three) ( rectangles) below the t-axis. Instead, the step function approximates the actual velocity v, and the dišerence of the areas of the rectangles approximates the change in the car’s position:

e area of e area of ⎡ ⎤ ⎡ ⎤ − ≈ ⎢ all rectangles ⎥ − ⎢ all rectangles ⎥ (Ô) s b s a ⎢ ⎥ ⎢ ⎥ . ( ) ( ) ⎢ above the t-axis ⎥ ⎢ below the t-axis.⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ Formula (1) would give a better approximation if we used more, narrower rectangles, as in Figure òÔ. To obtain exact results, we let the number of rectangles tend to innity and their widths tend to zero. We can expect that the areas of the two sets of rectangles determined by the graph of the continuous function v = v t would approach the areas of regions A and B in Figure ÔÀ, and that the change in position with velocity( ) given by the step function would approach the change in position of the truck with velocity given by v t . Consequently, we can expect that ( ) e area of e area of ⎡ ⎤ ⎡ ⎤ ⎪⎧⎢ ⎥ ⎢ ⎥⎪⎫ − = ⎪⎢ all rectangles ⎥ − ⎢ the rectangles ⎥⎪ s b s a lim ⎪⎢ ⎥ ⎢ ⎥⎪ ( ) ( ) ⎨⎢ ⎥ ⎢ ⎥⎬ ⎪⎢ above the t-axis⎥ ⎢ below the t-axis⎥⎪ (ò) ⎪⎢ ⎥ ⎢ ⎥⎪ ⎩⎪⎣ ⎦ ⎣ ⎦⎭⎪ e area of region e area of region = −  A in Figure òý  B in Figure òý where the limit is taken as the number of rectangles in the approximation tends to ∞ and their widths tend to ý. Lecture Ôý (Þ/ÔÀ/ÔÔ), p. À Math òýA

v v = v(t)

b a t

FIGURE òÔ

b In the next lecture we will dene the †•±u§Z v t dt so that it equals the dišerence of areas on ∫a ( ) the right of (2), and we will see that with this denition, equation (2) is an example of Part Ô of the F¶•oZ“u•±Z T„u™§u“ ™€ CZh¶¶«.

Interactive Examples Work the following Interactive Examples on the class web page, http//www.math.ucsd.edu/∽ ashenk/ (e chapter and section numbers on this site do not match those in the textbook for the class.) Section .Ô: Ô–À Section â.Ô: Ô, ò