DemysBCSolutionA-I.pdf - A Lu2019Hopitalu2019s Rule What...

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© 2011 - 7 - Demystifying the BC Calculus Exam A. L’Hopital’s Rule What you are finding: L’Hopital’s Rule is used to find limits of the form lim x ! c f x ( ) g x ( ) where lim f x ( ) = x ! c lim g x ( ) = 0 x ! c or lim f x ( ) = x ! c lim g x ( ) = " x ! c . How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or ! ! , apply L’Hopital’s rule, which says that lim x ! c f x ( ) g x ( ) = lim x ! c " f x ( ) " g x ( ) . L’Hopital’s rule can be applied whenever plugging in creates an indeterminate form: 0 0 , ! ! ,0 " ! , ! # ! ,1 ! ,0 0 , and ! 0 . A limit involving 0 ! " or " # " is found by creating a quotient out of that expression. A limit involving exponents ( 1 ! ,0 0 , or ! 0 ) involves taking a natural log of the expression to “move the exponent down.” Example 1: Find lim x ! 0 sin3 " x 2 " x lim x ! 0 sin3 " x 2 " x = 0 0 lim x ! 0 3 " cos3 " x 2 " = lim x ! 0 3cos3 " x 2 = 3 2 Example 2: Find lim x ! 0 + x 2 ln x lim x ! 0 + x 2 ln x = 0 "#$ lim x ! 0 + ln x 1 x 2 = #$ $ = lim x ! 0 + 1 x # 2 x 3 = lim x ! 0 + # x 2 2 = 0 Example 3: Consider the differential equation dy dx = 2 xy . Let y = f x ( ) be the particular solution to this differential equation with initial condition f 4 ( ) = 2. Find lim x ! 4 x " f x ( ) [ ] 2 sin # x ( ) . lim x ! 4 x " f x ( ) [ ] 2 sin # x ( ) = 4 " 2 2 sin4 # = 0 0 lim x ! 4 x " f x ( ) [ ] 2 sin # x ( ) = lim x ! 4 1 " 2 f x ( ) $ f x ( ) # cos # x ( ) = 1 " 2 f 4 ( ) $ f 4 ( ) # cos 4 # ( ) = 1 " 2 2 ( ) 2 4 ( ) 2 ( ) # = " 15 # There is no need to solve the DEQ. Example 4: The population of a bacterial colony is measured in thousands and is given by P t ( ) = 10 + t sin 1 t ! " # $ % & . Find the maximum limits of the population of the colony. lim t !" P t ( ) = 10 + "# 0 lim t !" P t ( ) = 10 + sin 1 t $ % & 1 t = 10 + ( 1 t 2 $ % & cos 1 t $ % & ( 1 t 2 = 10 + cos0 = 11 thousand ( )
© 2011 - 8 - Demystifying the BC Calculus Exam Example 5: A particle moves in the xy -plane so that the position of the particle at any time t is given by x t ( ) = e t ! t 2 ! t ! 1 and y t ( ) = t ln t ! t . Find lim t !" dy dx . dy dx = dy dt dx dt = t 1 t ! " # $ + ln t % 1 e t % 2 t % 1 = ln t e t % 2 t % 1 lim t &’ dy dx = lim t &’ dy dx = lim t &’ 1 t e t % 2 = lim t &’ 1 t e t % 2 ( ) = 0 Example 6: The figure to the right shows the graph of f x ( ) = x 2 . There is a horizontal line joining the coordinates x , x 2 ( ) and ! x , x 2 ( ) . Find the ratio of the area of the shaded triangle and the white area between the triangle and f x ( ) as x approaches zero. Area of Shaded triangle : 1 2 2 x ( ) x 2 ( ) = x 3 Area of white section:2 xt ! t 2 ( ) 0 x " dt = 2 xt 2 2 ! t 3 3 # $ % & ( 0 x = x 3 ! 2 x 3 3 = x 3 3 Ratio = lim x ) 0 x 3 x 3 3 = lim x ) 0 3 x 2 x 2 = 3 Example 7: Show that the area of the region bounded by f x ( ) = ! 2ln x and the x- axis for 0 < x 1 is finite. A = ! 2ln x ( ) dx 0 1 " = ! 2 x ln x ! x ( ) 0 1 = lim a # 0 + ! 2 x ln x ! x ( ) a 1 = lim a # 0 + ! 2 0 ! 1 ! a ln a ! a ( ) [ ] = lim a # 0 + 2 ! a + a ln a = 2 + lim a # 0 + a ln a = 2 + lim a # 0 + ln a 1 a = 2 + lim a # 0 + 1 a ! 1 a 2 = 2 + lim a # 0 + a = 2 This problem involves both improper integrals and integration by parts. A problem this involved has not been on the AP exam for quite awhile. But the concepts are fair game.