2.5 Evaluating Limits Algebraically
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1 SECTION.5 Evaluating Limits Algebraically 3.5 Evaluating Limits Algebraically Preinary Questions. Wic of te following is indeterminate at x? x C x ; x x C ; x x C 3 ; x C x C 3 At x, x isofteform 0 xc3 0 ; ence, tis function is indeterminate. None of te remaining functions is indeterminate at x : x C x and x C are undefined because te denominator is zero but te numerator is not, wile x xc3 xc is equalto0.. Give counterexamles to sow tat tese statements are false: (a) If f.c/is indeterminate, ten te rigt- and left-and its as x! c are not equal. (b) If f.x/exists, ten f.c/is not indeterminate. x!c (c) Iff.x/isundefined at x c, tenf.x/as an indeterminate form at x c. (a) Let f.x/ x x. At x, f is indeterminate of te form 0 0 but (b) Again, let f.x/ x x. Ten but f./is indeterminate of te form 0 0. x x! x.x C / x!.x C / x!c x!c x f.x/ x! x! x.x C / x! x x : (c) Let f.x/ x. Ten f is undefined at x 0 but does not ave an indeterminate form at x Te metod for evaluating its discussed in tis section is sometimes called simlify and lug in. Exlain ow it actually relies on te roerty of continuity. If f is continuous at x c, ten,bydefinition, x!c f.x/ f.c/;inoterwords,teitofacontinuous function at x c is te value of te function at x c. Te simlifyandlug-in strategyisbasedonsimlifyingafunction wic is indeterminate to a continuous function. Once te simlification as been made, te it of te remaining continuous function is obtained by evaluation. Exercises InExercises 4,sowtatteitleadstoanindeterminateform.Tencarryouttetwo-sterocedure:Transformtefunction algebraically and evaluate using continuity. x 36. x!6 x 6 Wen we substitute x 6 into x 36 x 6, we obtain te indeterminate form 0 0. Uon factoring te numerator and simlifying, we find 9.!3 3 x 36 x!6 x 6.x 6/.x C 6/ x!6 x 6.x C 6/ : x!6 Wen we substitute 3 into 9 3, we obtain te indeterminate form 0 0. Uon factoring te denominator and simlifying, we find 9!3 3.3 /.3 C /.3 C / 6:!3 3!3
2 4 C HAPTER LIMITS x C x C 3. x! x C Wen we substitute x into x CxC xc, we obtain te indeterminate form 0 0. Uon factoring te numerator and simlifying, we find x C x C x! x C.x C / x C.x C / 0: x! x! t 8 4. t!9 5t 45 Wen we substitute t 9 into t 8 5t 45,weobtainteindeterminateform 0 0. Uon dividing out te common factor of t 9 from bot te numerator and denominator, we find t 8 t!9 5t 45.t 9/ t!9 5.t 9/ t!9 5 5 : InExercises5 34,evaluateteit,ifitexists.Ifnot,determineweterteone-sideditsexist(finiteorinfinite). 5. x!7 x 7 x 49 x!7 x 7 x 49 x!7 x x!8 x 9 x 64 x!8 x x C 3x C 7. x! x C x C 3x C x! x C x! x 3 64x 8. x!8 x 8 x 7.x 7/.x C 7/ x!7.x C /.x C / x C x C 7 4..x C /. x! x 3 64x x.x 8/.x C 8/ x.x C 8/ 8.6/ 8: x!8 x 8 x!8 x 8 x!8 x 9x 5 9. x!5 x 5 x!5. C / 3 0. x 9x 5 x 5 x C. x! x C 3x C x!.x 5/.x C / x!5.x 5/.x C 5/ x C x!5 x C 5 0 :. C / 3 C 3 C 3 C 3 3 C 3 C 3 x C x C 3x C x!.3 C 3 C / 3 C 3.0/ C 0 3: x C.x C /.x C / x! x C : x x. x!3 x 9 As x! 3, tenumeratorx x! 6 wile te denominator x 9! 0; tus,tisitdoesnotexist.cecking te one-sided its, we find wile x x x!3 x 9 x x x!3c x 9 x!3 x!3c x.x /.x 3/.x C 3/ x.x /.x 3/.x C 3/ :
3 3x 4 3. x! x 8.3 C / t!0 4 t 4 t. C / 9 6.! x!6 3x 4.3x C /.x / x! x 8 x!.x /.x C / 3x C x!.x C / 8 8. SECTION.5 Evaluating Limits Algebraically 5.3 C / C 7 C 9 C C 9 C 3.7 C 9 C / 7 C 9.0/ C 0 7: 4 t tto0 4 t.4 t /.4 t C / tto0 4 t.4 t C / : t!0!4 x 6 x!6 t C 4 8. t! 3t t! y C y 9. y!3 y 3 0y C 3 y!3. C / C / C 4. /. 4/!4 4!4 4 x 6 x!6 x C 4 x!6 t C 4 3t t! y C y y 3 0y C 3 y!3.c/ 4.t C / 3.t /.t C / t!.y 3/.y C 4/.y 3/.y C 3y / y!3 4.C/ 4.C/ 4 4.C/ x C 4 8 : 3.t / 6. C. C does not exist. C C. C C / As! 0C, weave. C C / C C. C C / As! 0,weave. C C /. x!8 x 8. / 3:!4.y C 4/.y C 3y / 7 7 : 4. C4C4/ 4.C/ 4 4.C/ 4 4. C / :. C C /!.. C C /!.
4 6 C HAPTER LIMITS x!8 3. x 8 x 4. 5 x x x 8. /. C / x!8.x 8/. C / x!8 x!8 x 8 x C 4 C 4 : 4.x 8/. C /./. x C 8 x/. x 8 x/. x C 8 x/./. x C 8 x/ x.8 x/./. x C 8 x/./. x C 8 x/ x 8./. x C 8 x/ 4 C 4 : 5 x x! 5 x 5 x C 4 x x 5 x C. x/. 5 x C / 5. 4 x 4 x 6. x x!0c x C x x x!0c x C x. x/. C x/. x/. 5 x C / x!0c x!0c cot x 7. x!0 csc x cot x x!0 csc x cos x sin x cos 0. x!0 sin x 8.! cot csc! 9. t! t C t 0 t 4 cot csc! t C t 0 t! t x! x x x C 4 x x C cos sin sin cos 0. C x 5 x C : x x x C 4 : x C x C x C C x x C x!0c x x C x C C x x x C x C C x!0c. t C 5/. t 4/ t! t. t C 5/ 9: 4 t! x x C x C C 0:
5 sin x cos x 3. x! tan x 4 3.! 33.! 4 x! x x x! sin x cos x x! tan x 4 sec tan sin sec tan!! cos tan! 4 tan tan tan cos x C 3 cos x 34. x! cos x 3. C x/. x/. C x/ x! SECTION.5 Evaluating Limits Algebraically 7 C x : cos x cos x.sin x cos x/cos x cos x! sin x cos x 4. 4! 4 cos x C 3 cos x x! cos x 3 x! Use a lot of f.x/ algebraically in Exercise 3. C sin C sin! sin cos. C sin /!.tan C /.tan C /.tan /! 4. cos x /.cos x C / cos x tan C : cos C sin 0 0: cos x C cos x! 3 C 5 : 3 toestimate f.x/to two decimal laces. Comare wit te answer obtained x 8 x x 4 Let f.x/. From te lot of f.x/sown below, we estimate f.x/ :00; totwodecimallaces, x 8 x tis matces te value of obtained in Exercise Use a lot of f.x/ 4 toestimate f.x/numerically. Comare wit te answer obtained alge- x braically in Exercise y x Let f.x/ 4 x x 4. From te lot of f.x/ sown below, we estimate f.x/ 0:5; totwodecimal laces, tis matces te value of 4 obtained in Exercise 5. y x In Exercises 37 4, evaluate using te identity a 3 b 3.a b/.a C ab C b /
6 8 C HAPTER LIMITS x x! x x 3 8 x! x.x / x C x C 4 x C x C 4. x! x x! x x!3 x 9 x 5x C x! x 3 x 3 C x! x C 6x C 8 x 3 7 x!3 x 9.x 3/ x C 3x C 9 x!3.x 3/.x C 3/ x!3 x 5x C 4 x! x 3 x! x! x 4 4. x! x 3 4. x!7 x 3 C 8 x C 6x C 8 x!.x /./.x / x C x C x!.x C /.x x C 4/.x C /.x C 4/ x C 3x C 9 x C x C x C. x!.x x C 4/ x C 4 6: x 4 x! x 3.x /.x C / x!.x /.x C x C /.x /.x C /.x C /.x C /.x C / x!.x /.x C x C / x!.x 4 C x C / 3 : x 7 x =3 3 x!7 43. Evaluate 44. Evaluate x 7 x =3 3.x =3 3/.x =3 C 3x =3 C 9/ x!7 x =3 3 x!7.x=3 C 3x =3 C 9/ 7 4 C. Hint: Set x 4 C and rewrite as a it as x!. Let x 4 C. Ten x 4.x /.x C /.x C /, x! as! 0 and 4 C x! x.x /.x C /.x C / x! 3 C. Hint: Set x 6 C and rewrite as a it as x!. C.x C /.x C / 4 : Let x 6 C. Ten 3 C x.x /.x C /, C x 3.x /.x C x C /, x! as! 0 and 3 C C x!.x /.x C /.x /.x C x C / x! x C x C x C 3 : InExercises45 54,evaluateintermsofteconstanta. 45..a C x/ x! a C 7a/!.a C x/ a. x!0.4a C 7a/ a.! 47..4t at C 3a/ t!.4t at C 3a/ 4C5a. t!.3a C / 9a 48.
7 .3a C / 9a 6a C.a C / a 49..x C a/ 4x 50..a C / a 4a C x a 5. x a a C a 5..x C a/ 3 a x!0 x 54.!a a a.6a C / 6a. SECTION.5 Evaluating Limits Algebraically 9.4a C / 4a..x C a/ 4x.x C ax C a / 4x 3x C ax C a x!a a C a.a x/.a C 3x/ x a x a x C a x!a..a C 3x// 4a: x!a x C a a. a C a a C C a a C C a a C C a : a C C a a.x C a/ 3 a 3 x 3 C 3x a C 3xa C a 3 a 3 x!0 x x!0 x x!0.x C 3xa C 3a / 3a : a!a a!a a a a a!a a Furter Insigts and Callenges a!a InExercises55 58,findallvaluesofc suc tat te it exists. a a x 5x x!c x c x 5x 6 will exist rovided tat x c is a factor of te numerator. (Oterwise tere will be an infinite x!c x c discontinuity at x c.) Since x 5x 6.x C /.x 6/,tisoccursforc and c 6. x C 3x C c 56. x! x x C 3x C c exists as long as.x / is a factor of x C 3x C c. Ifx C 3x C c.x /.x C q/, ten x! x q 3 and q c. Henceq 4 and c x! x c x 3 Simlifying, we find x c x 3 x C x C c.x /.x C x C / : In order for te it to exist as x!, tenumeratormustevaluateto0atx. Tus,wemustave3 c 0, wicimlies c 3.
8 30 C HAPTER LIMITS C cx C x 58. x!0 x 4 Rationalizing te numerator, we find C cx C x x 4. C cx C x /. C cx C C x / x 4. C cx C. C cx /. C x / C x / x 4. C cx C C x /.c /x C c x 4 x 4. C cx C C x / : In orderforteittoexistasx! 0, tecoefficientofx in te numerator must be zero. Tus, we need c 0, wic imlies c. 59. For wic sign does te following it exist? x!0 x x.x / Te it x!0 x C x.x / Te it x!0 x x.x /.x / C x!0 x.x / x!0 does not exist. x. As x! 0C, weave x.x / x x.x / x.x / x.x /!. As x! 0,weave x.x / x x.x / x.x / x.x /!..6 Trigonometric Limits Preinary Questions. Assume tat x 4 f.x/ x.watis f.x/?istereenouginformationtoevaluate f.x/? Exlain. x!0 x! Since x!0 x 4 x!0 x 0,tesqueezeteoremguaranteestat x!0 f.x/ 0.Since x! x x! x, we do not ave enoug information to determine x! f.x/.. State te Squeeze Teorem carefully. Assume tat for x c (in some oen interval containing c), and tat l.x/ u.x/ L. Ten f.x/exists and x!c x!c x!c 3. If you want to evaluate l.x/ f.x/ u.x/ f.x/ L: x!c sin 5, it is a good idea to rewrite te it in terms of te variable (coose one): 3 (a) 5 (b) 3 (c) 5 3 To matc te given it to te attern of sin!0 ; itisbesttosubstituteforteargumentoftesinefunction;tus,rewriteteitintermsof(a): 5.
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