Calc 2 Chapter 5.8 and chapter 6 intro
Calc 2 Chapter 5.8 and chapter 6 intro MTH 162
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This 6 page Class Notes was uploaded by Jack Magann on Tuesday February 10, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 140 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.
Reviews for Calc 2 Chapter 5.8 and chapter 6 intro
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Date Created: 02/10/15
Calculus 2 Chapter 58 In previous chapters we were unable to solve such equations as this 62quot 63quot 2 lim x gt0 3 sm2x 2 COSCX 2 Since using the substitution method shows the bottom of the equation to be 0 it s limit would be DNE Does Not Exist However using L Hopital s rule this limit can be solved L Hopital s rule Let f x and gx be differentiable functions If 1 limxa f x limxa gx 0 or 2 limx m f x limx m 906 too Then f x 9 06 limxa imxa if both limits exist gx We can now solve the limits we were not able to in earlier chapters Ex 2x 3x e e 2 o 1 11mx0 smce the 11m1t of both equatlons equals zero we can 3 Sln2x 2 cosx2 0 use L Hopital s rule f x 51 96er 63quot 2 262x 363x g x 2 3 sin2x 2 cosx 2 6 cos2x 2 sinx 2e2x3e3x 23 5 11mx0 Now the subst1tut10n techmque prov1des a real 6 cos2x2 Slnx 60 6 answer Certain problems will require that you make use of L Hopital s rule multiple times EX 1 lim w 3 Now we can use L Hopital s rule x 3x32x 1 oo 6x2 10x oo llmx mo m Slnce 1t stlll equals 00 we use L Hopltal s rule aga1n 12x 10 oo 11mxoo 18x 2 Slnce 1t stlll equals 00 we use L Hop1tal s rule aga1n um 12 2 Xquot 18 18 00 and g are called indeterminate forms however they are not the only ones 0 OIO m Indetermmate Forms 5 0X00 oo oo 00 1 00 What s important to know is that L Hopital s rule only works for 00 and so you have to change the other indeterminate forms into them Ex exe x 2 o 1 11mx0 m 0 Use L Hopltal s rule lim ex e x 0 Use L Ho ital s rulea ain lim exe x 1 9 60 25in2x 0 p g 9 60 4cos2x 2 lntanx oo oo 2 11m x60 1ntan 2x oo 00 SEC2 x lirr1 limx0 cotx sec2 x tan 2x cos2 2x 95 tan 2x Use sin2x 2 sin x cos x so when you cancel everything out you get cos2 x 1 lim 1 9 60 cosx 1 3lim ln2x35x21 oo lim 6x210x 3x47x2 xquot 1n3x47x2 xquot 2x35x21 12x37 Rather then multiplying these expressions out we can just nd the values of the highest degrees of X The limit is then just the fraction that the numbers in front of the highest degrees of X make so 6x210x 3x47x2 18x6 18 lim 11m xquot 2x35x21 12x37 xquot 24x6 24 4 lim 1 1 lim x 0 x gt0 ex1 x x 0 xex x 0 1 ex 0 ex 1 11m gt11m 9 60 xexex 1 0 9 60 xexexex 2 sin 2x 0 5 11m sm 2x csc 3x x60 sin 3x 0 2cos2x 2 lim x 3 cos 3x 3 For the indeterminate forms 1 00 000 let L lim fxgx lnL lnlim Uxgx lim 1anxgx 1 lnL 2 11m gx lnfx 2 11m n11 m 1nfx 1 lim lnL 61 e 906 Ex 1 1 limx0 6quot 3x 10 L 1 lnex3x 4 x so lnL 11mx01nex 3xx 11mx0 9 1393quot 2 E 4 This however is not the answer according to the equation above The full answer is e4 Techniques of Integration Chater 60 1 Division For when there are two polynomials in the fraction and the highest degree of X in the numerator is equal to or greater then that of the bottom Also the equation can not be broken into parts 2 f2x 11 x24 2x2x11 2x2 8 x 3 2x2x11 x 3 f x24 f2 x24 x 3 1 2 3 1 x f2fxz8 fx24 2x lnx 8 tan C 2 Rationalizing Substitution fijr J lett Vx 1 nowputxanddxinterms ofy xt2 1 dx2tdt f f tht t212dtf2t22dt t3 21 vx 13 2 mc 3 Completing the Square f x2x613 dx If the bottom polynomial was in the form of x2 bx or x2 bx the integral could easily be done To do this Beak the last number up for bottom to be perfect square plus or minus removed number x1 x1 dx fx2 6x13 fx 324 Obtain the form x2 bx b2 or x2 bx b2 Now use substitution using the completed square wx 3 dwdx and xw3 f x1 x fW31 x 324 w24 W 4 l 2 1K fwz4fw24 21nw 42tan 2 1nx 6x 13 2tan 1 C All these notes are from a Calculus 2 class run by Dr Patrick Dibby at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print
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