Unformatted text preview: 2.1 RATE OF CHANGE AND LIMITS e right, g(x) approaches 0. As x approaches 1 from the left, g(x) L that all the values g(x) get arbitrarily close to as x Ä 1. 2 (b) 1 CHAPTER c) 0 LIMITS AND CONTINUITY Thomas’ Calculus 11th Edition, C(hapter 2, Section 1: 1, 3, 5, 11a, 13a, 17a, 29, 31. 2.1 Total: 8 CHANGE AND LIMITS RATE OF Mathematics I: Homework 7. The tangent and velocity problems. that all the values g(x) get arbitrarily approaches 1. There is no single number L Limits. 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approa 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) (c) Does not exist. As t approaches 0 from the left, f(t) approaches c1. As t approac approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. 2.1 c RATEapproaches 0 AND LIMITSf(t) RATE t OF e left, f(t) approaches2.11. 1As OF CHANGE from the LIMITS approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä (b) Answers: CHANGE AND right, L that f(t) gets arbitrarily close to as t Ä 0. (c) 0 2. (a) CHAPTER 20 LIMITS AND CONTINUITY CHAPTER c1 LIMITS AND CONTINUITY 2 (b) (b) (e) (b) (e) x 0 and e right, Ä 0.
x kx k 1. 1(a)(a) Does not exist. Asapproaches .1 fromTrueright, g(x) approaches 0. 0. Asapproaches 1 from the left, g(x) Does not exist. As x x approaches (a)from the right, g(x) approaches As x x approaches 1 from the left, g(x) 3 1 the (b) True . 1. approaches 1. 1. There no no single(False L that allall the values g(x) get arbitrarily close to as x Ä 1. There is is single number L that the values g(x) get arbitrarily close to as x Ä 1. d) False (e) False True 2. (a) 0 approaches (c) number (b) (b)1 1 b) c 1 ( False (f) True (c) (c) 0 not exist. As t approaches 0 from the left, f(t) approaches c1. As t approaches 0 from the right, f(t) (c) Does 0 4. (a) False (b) False approaches 1. There is no single number(L that f(t) gets arbitrarily close to as t Ä 0. d) True (e) True False (c) True 3. 0 2. 2(a)(a) 0 . True 3. (a) True (b) True (c) False x (b)(b) 1c1 c 5. lim kxk does not exist because kxk œ x œ 1 if x 0 and kxk œ cx œ c1 if x 0. As x x x xÄ0 x (d) False (e) False (f) True x x (c)(ifDoes 0. not x approaches 0 from 0 from thethe left, f(t) approaches1. 1. As t approaches from the right, f(t) not exist. As t approaches the from t x c)x Does As exist. As t approaches 0xleft, left, f(t) approaches c c As 0 approaches 0 0 from the right,1. There is no si f(t) k x k œ cx œ c 1 kxk approaches c1. As x approaches from the right, kxk approaches approaches 1. 1. There no no single number that f(t) gets arbitrarily close to to as ÄÄ 0.(c) True There is is single number L LFalse f(t) gets arbitrarily close as t t 0. a 4. 1. Falsepproaches (b) that approaches (a) There is no single number L that all the function values get arbitrarily close to as x Ä 0. 5. (d) True (e) True 3. 3(a)(a) True True (b)(b) True True (c)(c) False False . . x from the left, values of x " 1 become increasingly large 11a. x 6 not exist because x œ x œ61 ifAs 0approachesx 1œ c1 if x 0. the x approachesc0 from the left, (f) True and nega
does x 5. (d)(d)xk False lim False x and kxk (e)(e) False œ cxFalse As (f) True kx k x xÄ0 k " f As x approaches 1rom the x xc1 become increasingly large and negative. approaches 0 from the right, xright, the values become increasingly large and positive. There is no one num 13a.
1/8 c1. As x kxk approaches kxk approaches 1. There is no single number L that all " function values get arbitrarily close to as x Ä 1, so lim xc1 does not exist. large and positive.4.There is no one number L that all the False (b)(b) False False (c)(c) True True 4t(a)(a) False .he function values get arbitrarily close to as x Ä 0. xÄ1 " 1, so lim xc1 does not (d)1 True (d)exist. True (e)(e) True True 17a. xÄ1 " 7 xc1 become be said about f(x) because the existence of 1 6. As x approaches 1 from the left, the values. ofNothing canincreasingly large and negative. As x approachesa limit as x Ä x! does not de xxxx x 29. limxk right, the values on becausek functionœif x positive.xkTherexxisfor œ 1 if xto exist, f(x)approaches 0 from the left, a single r 19 does notnot exist how increasinglyslarge and at and In order xnoclimitif 0. As xthe be arbitrarily closeleft, f.lim kx stence of a limit as 5. 5xrom0the kxk not depend become the kœ œ œ 1 1 if x x! . kx kœ œ cx a cnumber L that x must x ÄÄx! does does exist because kx xk x i defined 0 0 and xk c œ one 1 x 0. As all approaches 0 from the to x x xÄ0 " function a single real number L when arbitrarily close to as 1, so x x to values x is close enoughapproaches 1. There is is single number L that allall to x! . That x) must be arbitrarily closepproachesget 1. 1. Asapproachesx0 Ä from xlim1 xc1xdoesapproaches 1. There nono of a limit depends on the values of f(x) x not exist. is, the existence a c right, kx 0 the single number L that kxk k approaches c As x x approachesfrom the Ä right,k kxk 31. x
k4/π of a limit depends on the values of f(x) for x near x! , not on thedefinition of f(x) at x! itself. thethe function values get arbitrarily close as as Ä Ä 0. function values get arbitrarily close to to x x 0. 7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily closeorder for lim number L whenf(x) must close to a single value f 8. Nothing can be said. In to a single real f(x) to exist, " 6. 6x isAsapproaches 1x!1 That is,left,existence of a limit1depends on the values of large andnearnegative. the approaches 1 1 As x x approachesfrom thethe left, the values of xc1 become increasingly x Ä 0xand x! , not on As x approaches become increasingly f(x) for negative. As x . close enough to . from the the values of xc " large exist, f(x) must close to a single value for x near 0 regardless ofhe value f(0) itself. t fefinition of right, x!values become increasingly large and positive. There is is one number L L that the drom thethe f(x) thethe values become increasingly large and positive. There nono one number that allall the from right, at itself. " function values getget arbitrarily close as as Ä Ä so so lim c1 c1 does not exist. function values arbitrarily close to to x x 1, 1, lim x x "does not exist. x Ä Ä 1 not require that f be defined at x œ 1 in order for a limiting val x 1doesa single value for x near 0 regardless of 9. No, the definition 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to xÄ0 i fined at x œ 1 in order tforvalue f(0) itself. to exist there. If f(1)s defined, it can be any real number, so we can conclude nothing about f(1) from lim a limiting value he xÄ 7. 7Nothing can be be said about f(x) because the existence oflimit as as x Ä ! x! does not depend on how the function . Nothing can said about f(x) because the existence of a a limit x Ä x does not depend on how the function n conclude nothing about f(1) from lim f(x) œ 5. is defined at x ! .x! . In order forlimit to exist, f(x) must be be arbitrarily close tosingle real number L when x Ä 1 order for a a limit to exist, f(x) must arbitrarily close to a a single real number L when In is defined at 9. No, the definition does not require that f10.defined at x œ 1the order for a limiting value to exist there. values of f(x) when x is near 1, no be No, because in existence of a limit If f(1) x defined, it can be anyxreal!number,is, the can conclude limit depends onon thelimdepends5. forfornear x! ,xnot onon the is close enough to to .x That is, thewe existence ofnothing about f(1) fromvalues of of f(x) the x near ! , not the existence of a a limit depends the values œ on x f(x) is x is close enough ! . That so f(x) n the values of f(x) when xefinition f(x) at x f(1) itself. If x some lim f(x) exists, its value may be Ä 1 number other than f(1) œ 5. We can conclude definition of of f(x) at x! itself. d is near 1, not on ! itself. xÄ1 r other than f(1) œ 5. We can conclude nothing about lim f(x),hether it exists or what its value is if it does exist, from knowing the value of f(1) alo xÄ1 w 10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If xist, from knowing . 8Nothing exists,alone. In In order for lim f(x) exist,f(1)f(x) must can conclude nothing aboutx limnear regardless of of f(1) be said. order for number other than f(x) 5. We close a a single value 8 the.lvalue ofcan beits value may be some lim f(x) to to exist, œmust close to tosingle value forfornearf(x), regardless said. Nothing can x 00 im f(x) thethe value f(0) itself. value f(0) or what whether it existsitself. its value is if it does exist, from knowing the value of f(1) alone. xÄ1 xÄ0 0 xÄ xÄ1 9. 9No, thethe definition does not require that f be defined x œ œin in order forlimiting value to to exist there. f(1) . No, definition does not require that f be defined at at x 1 1 order for a a limiting value exist there. If If f(1) is defined, it can be be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. is defined, it can any real number, so we can conclude nothing about f(1) from lim f(x) œ 5.
xÄÄ1 x1 10.10. No, because the existence oflimit depends on on the values of f(x) whenis near 1, 1, not on f(1) itself. If No, because the existence of a a limit depends the values of f(x) when x x is near not on f(1) itself. If lim im f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), l f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x),
xÄ1 1 xÄ x ÄÄ 1 x1 whether it exists or or what its value if it does exist, from knowing the value of of f(1) alone. whether it exists what its value is is if it does exist, from knowing the value f(1) alone. [email protected] ...
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This note was uploaded on 12/26/2010 for the course MA 1012 taught by Professor Franciscoperez during the Fall '10 term at ITESM.
 Fall '10
 FranciscoPerez
 Math, Continuity, Rate Of Change, Limits

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