egy that would eventually work), but these concepts were763 not well Heine understood until 6965_CH10_pp735-778.qxd 1/20/10 3:35 PMhis Page Karl (1815–1897) and student Eduard (1821–1881) introduced Karl Weierstrass (1815–1897) andWeierstrass his student Eduard Heine (1821–1881) introduced Karl Weierstrass (1815–1897) and his student Eduard Heine (1821–1881) introduced the formal, unassailable definitions that are used in ourtohigher mathematics courses tothe formal, unassailable definitions that are usedPage in our higher mathematics courses 6965_CH10_pp735-778.qxd 3:35 PM the formal, 1/20/10 unassailable definitions that763 are used in our higher mathematics courses today. By that time, Newton and Leibniz had been dead for over 150 years. day. By that time, Newton and Leibniz had been dead for over 150 years. day. By that time, Newton and Leibniz had been dead for over 150 years. Excelsior PreCalc Honors Name____________________________ 12.1 – Evaluating Limits Numerically by Substitution 6965_CH10_pp735-778.qxd PM 763 Page 763 6965_CH10_pp735-778.qxd 1/20/10 1/20/10 3:35 PM3:35 Page a Limit Informally DefiningDefining a LimitDefining Informally a Limit What do you recall about the limitInformally of each of the following functions: S Warm up There nothing difficult about limit the following limit statements: There is nothing difficult about the is following statements: There is nothing difficult aboutlimit the following statements: SECTION 10.3 More o 1 1 2 12x - 12q= 25 lim 1x 20 + lim 32 =1 q lim = 0 lim = 5 12x -lim 1x + 32 lim = 1x + lim = lim 12x - 12 lim q 12 = 5 32 = = 0 q q x:3 q n n: q n x:3 x:SECTION n:x: EXERCISES x:3 x : q 10.3 n: q n SECTION 10.3 EXERCISES That is why have usedbackground limit notation throughout this Particularly when That is why we have used limit notation throughout this book. Particularly when That is why we have usedwe limit throughout this book. Particularly Exercise numbers with anotation gray indicate problems thatbook.when 29. (a) lim ƒ1x2 x : 3graphers the limiting behavioralgeof functions algeelectronic graphers are available, analyzing the are limiting behavior of functions algetheelectronic authors designed toavailable, be solved without abehavior calculator. electronic graphers arehave available, analyzing the analyzing limiting of functions numbers with a graysubstitution: background indicate problems that 29.much (a) lim ƒ1x2 we need ƒ1x2 y (b)to lim Example 1 Exercise Find the limitand by direct braically, numerically, and graphically canneed tellifwhat us of -what know braically, numerically, graphically can telltheuslimit much of we to know braically, numerically, and graphically can tellwhat ussubstitution much of we need to know x:3 x: 3+ In Exercises 1–10, find by direct it exists. the authors have designed to be solved without a calculator. about the functions.2 about the functions. about the functions. (b) lim+ ƒ1x2 (c) lim ƒ1x2 12 lim 1x -substitution 12 2. lim 1x - 12 In Exercises 1–10, find the1.limit by xdirect if b. it exists. a. x : -1 x:3 x:3 x:3 4 3 What difficult is to with an air-tight definition what a(c) limit If really is. If 2 ƒ1x2 What is come difficult is to come upofwith anaair-tight definition ofreally whatis. a limit What is difficult is toiscome up with an air-tight definition what limitofreally is. If lim 3 up x:3 - 1212 1. lim x 1x - 122 3. lim 1x2. - lim 2x 1x + 32 4. lim 1x 3 - x + 52 it had been easy, itx : would not taken 150 years. subtleties of the “epsilonx: -1 x:3 it had been easy, ityears. would not have years. The subtleties of the “epsilonit had been easy, it would not have 150have The subtleties of150 the “epsilon2taken x :taken -2 The 1 3 3 2/3 (a) limbut ƒ1x20 lim 1x 2x + 32 lim 1x x + 52 3. 4. delta” definition of Weierstrass and Heine are as beautiful as they are profound, but lim lim 1x + are 5 ofasWeierstrass 1x -profound, 42 as beautiful 5. and delta” definition and Heine are profound, delta” definition of Weierstrass Heine beautiful 6. as they are but as they are30. 6965_CH10_pp735-778.qxd 1/20/10 Page x : 1x:2 x: -23:35 PM x:2 x : -2 763 they are not the stuff of acourse. precalculus course. Therefore, even as welimlook more they are not stuff precalculus even as course. wePMlook more are not of42a2/3 precalculus Therefore, even more 6965_CH10_pp735-778.qxd 3:35 Page (a) 763 ƒ1x2 as we look (b) lim ƒ1x2 y limstuff lim 1x +of5 a they 1x Therefore, -1/20/10 5. the 6. the x 30. + x properties x:1 x : 1 x:2 x: -2 closely at limits and their in this section, we will continue to refer to our lim lim 1e sin x2 ln a sin b 7. 8. Exercise numbers with a gray background indicate problems that closely at limits and their properties this section, will continue to section, refer to we our will continue to refer to our 29. closely atinlimits and theirweproperties in this x:0 x:p 2 We (b) limit ƒ1x2 ƒ1x2 4 (c) lim for x that “informal” definition of limit (essentially of d’Alembert). here for xthe “informal” 7. definition of limit (essentially that of d’Alembert). We repeat it here for “informal” definition of limit (essentially that of d’Alembert). x:1 + We repeat it here authors have designed without a repeat calculator. x:1 lim lnbackground lim 1e sin x2 asin to b be solved 8.gray Exercise numbers with a indicate problems that 2 29. (a)2 ready reference: x:0 x:p 2 x - 1 ready reference: ready reference: (c) lim ƒ1x2 9. lim 1x 2 - 22 10. lim 1 SECTION 10.3 EXERCISES SECTION 10.3 EXERCISES x:1 the authorsInhave designed be solved asubstitution Exercises find the limit without byx : direct if it exists. x : a1–10, to a x2 + 1calculator. x2 - 1 0 (b)the In Exercises 31 and 32, 9. lim 1x 2 - 22 10. lim 2 2 12 In1. Exercises 11–18, (a) explain why you cannot use substitution to find x:a x:a x by + 1direct substitution limsubstitution: x 1xlimit - 12 2. lim 1xif-it 12 Which of the statements ab InFind Exercises find the exists. You Try the limit1–10, by thedirect limit and limit at algebraically if itx:3 exists.In Exercises 31 and 32, the graph of a function y = x: -1 (b) find theLimit (INFORMAL) a DEFINITION (INFORMAL) Limit at a In ExercisesDEFINITION 11–18, (a) explain why you cannot use substitution to find DEFINITION (INFORMAL) Limit at a2 Which (c) lim are ƒ1x2 31.the (a)function true= 12 of the statements about x : -1 + x32 + = 7x 12 x 12 - 91x 3 - close lim 1x2ƒ1x2 2x 32mean lim x + to52L 3. 4. lim limfind xthe 1x 12 1x 1.andWhen 2. the limit (b) limit algebraically ifL+it,”+exists. lim we write “ we that ƒ1x2 gets arbitrarily lim lim 11. 12. a. b. L,” When we write x: “ lim we that ƒ1x2 gets=x:3 arbitrarily close to 31. Lƒ1x2(a) x:2 -2 ƒ1x2 L,” we mean write “ lim that getslim arbitrarily to L lim ƒ1x2 = -1ƒ1x2 =When x: awemean ƒ1x2 =close 1 (b) 2x: 2 x 2 -not 9x:a x: -1 + + 7x + 12 x : -3 x 2 - 9to a. x : 3 x 3+ 2x - 15 x : 0as xx:a gets arbitrarily close (but equal) 2/3 3 30. as x gets11. arbitrarily (but not equal) to a. lim limlimclose 12. lim lim 1x + 5 1x 42 5. 6. as x gets arbitrarily close (but not equal) to a. 3 3 2 1x 2x + 32 lim 1x x + 52 3. 4. 2 x +x:3 1 x 2 + 2x - 15 x x: - 2x + x -(b)2 lim- ƒ1x2 = 0 (c) lim- ƒ1x2 = 1 x: -3 x 9 x:2 -2 x:0 x:2 x: -2 x:0 lim 13. lim 14. x : -1 x + 1 x - 2/3 2 (c) lim ƒ1x2 = 1 x3 + 1 x 3 - 2x 2 + x - x2: 2 lim ƒ1x2 (d) x 30. (a)= x2 lim +lim 5 1e14. 6. lim 8.1xƒ xlim x:0 2- 42 13.5.limlim 1x x: 0sin x2 ln asin b 7. 4 ƒ x 4 x: -1 x:2 x + 1 x 2 x:2 x: -2 (d)2 lim- ƒ1x2 = lim+ ƒ1x2 x:0 lim lim x:p 15. 16. (e) lim ƒ1x2 exist 2 x:0 x:0 x : -2 x + 2 x : -2 x + 2 x:0 2 x 4 ƒ ƒ x - 4 (b)ƒ1 x x Exercise limbackground 15. lim numbers 16. lim= lim problems ƒ1x2 exists 29. (e)that lim (f) ƒ1x2 1 (g) 2 Exercise with a gray indicate problems numbers with a gray background indicate that 29 lim (a) ƒ1 lim 7. - 2 x b- 1 x:0 x: -2 x: -2 x + 28. lim lnx asin x:0 x + 1e 2 sin x2 x:0 2 - 3 lim lim 1x 17. 18. x:3 x:0 x:p 2 lim -have 22 solved lim 9.the 2 designed to x10. be solved without the authors have designed be without calculator. xauthors : 01x to : 0a lim ƒ1x2 = 1 (g)a calculator. ƒ1x2lim = ƒ1 0 (i) lim (h) (c) xx:a x - 2 x:a x:1 ƒ1 x(b) :1 x 2 + 1 x:0 lim 17. lim 1x - 3 18. lim + sin 2x direct x:0 x:0 x:3 lim ƒ1x2 = 0 32. (a) lim(j) ƒ1 (i) lim if Infind Exercises 1–10, limit by it exists. In Exer ƒ1x2 x 2find In Exercises 1–10, the limit by direct itsubstitution exists. -= 1, along lim x =cannot 1if In2 Exercises 19–22, use the fact the thatsubstitution withsubstitution the x:1 x : -1 + x:2 In Exercises 11–18, (a) explain why you use to find x : 0 x 9. lim 1x - 221.2 lim sin lim ƒ1x 2 lim 12 ƒ1x2 = 1 (b) (c) x - 10. Which 12 lim 32. lim lim ƒ1x2 x the 1x 1x (a) - 12 2 122. + lim = following 12. In Exercises 19–22, use fact12 that , 12 along with the x:3does lim lim xlimit 1xtheproperties, -and 1x 1. x:a x:a limit to find the limits. x: -1 the (b) find limit algebraically if it exists. x : 2 x + 1 x: -1 x x:3 x:0 x: -1 x:3 ƒ1x2 31. = 2 (c) Inxlim Exercises 3 3 lim ƒ1x2 does not exist sin x sin 3x limit properties, to find3 the following limits. 2 2 (b) 3 lim 1x7xwhy -+2x + lim 32 lim 1x -x:2 x to + find 52 3.+explain 4. :2 xx:2 + 12 x 9 1x 19. - 2x 32 1x x + 52 3. lim 11–18, 4. lim lim 20. In Exercises (a) you cannot use substitution x : 0 2x 2 - x :0 xlim x: -2 (c) lim ƒ1x2 = 2 Which of the x:2 lim ƒ1x2 = (d) sin x 11. lim sin 3x x: -2 x12. 2/3 2 2 1x - x:2 x : 1x:5. -3 20. x:3 30 the19. limit (b) find the limit algebraically if it exists. lim lim and 2/3 lim lim 1x + 5 42 6. 2 x 9 x + 2x 15 2 1x + 5 (a) ƒ1x2 lim-=ƒ11 x -2 (d) lim ƒ1x2 = 2 30. lim 1x 5. 2x c.x:0 d. -x +42sinx: - x 21. lim sin x:2x x:0 x 6. lim (e) lim 31. (a) x:2 x:22. -2 xlim x: 1 + x:1 3 3 2x:12 2 x:0 x x : 0 2 2x + 1x x + sin x x 9 - 2x + x 2 sin x x + 7x + 12 x x lim+bƒ1x2 = 1 limdoes ƒ1 lim ƒ1x2 (f ) (b) lim lim 13. lim 21. 22. lim 14. lim lim x:0 1e sin asinx:1 7. 23–26, 8.x lim ln (e) 11. 12.limlim x : 1 x:1 + find the2xx2 limits. x:0 x 1eInx Exercises lim sin x2 ln asin b 7. x: 8. x: -1 x:2 2 2 x + 1 x 2 x:0 x:p (b) 2 -3 x:3 x + 2x - 15 x - 9x (f) lim ƒ1x2 does not exist x:0 x:p (g) lim+ ƒ1x2 = 3 sin2x - 4 cos (c) In Exercises 23–26, find the limits. e2 - 1x x : 0 lim ƒ1x x 2 2 x x:1 23. lim x - 4 24. lim 3 x 4 ƒ ƒ x:1 3 2 x 1 lim ƒ1x2 = lim ƒ1x2 (g) :x0 5x + cos x (h) lim ƒ1x2(c) exist x + lim 1x : 0 log 2x 2+ 41x + 222 ex - 15. 1x x - 4 cos x x216. lim x:0 - 10. 1sin lim 1x 3 -sin22 lim +2 x -x:0 x:c 2 x:9. lim lim 13. 14. lim lim lim 23. 24. -2 x: -2 x:a x:a x + 2 x + 2 ln12x2 1x 22 lim 9. 10. 1x +x9 - x2 (h) + 1lim ƒ1x2 exists for every x:0x: x:0 5 sin x + cos log4-1 1x +x22+ 1 x 2 1-1, 12. c in ƒ1x2 (i) lim exists x:2 x:a x:a lim 25. lim 26. x:c x:c x + 1 In Exe (d) 2 x : p/2 x : 27 sin 11–18, x 1x (a) ln12x2 In Exercises 3 xcannot + 9explain why2log you uselim substitution to find33 x4 - 2 (i) Exercises 31 and ƒ1x2 exists 11, 32 for every c inand .use 2 InIn Exercises 34, lim lim 26. Which x x:c x 4 In25. Exercises 11–18, (a) explain why you cannot use substitution to find lim lim 1x 3 17. 18. 27–30, use the given graph to find the limits or to explain x:p/2 sin2 xIn Exercises x:27 log x the limit and (b) find316. the limit algebraically if it exists. Which ofof,the lim+ ƒ1x2 lim (c)(e) x:0 x:0 In Exercises lim lim 15. 33 and 34, (b) use ƒand to statem find (a) x2 why the limits do notalgebraically exist. xa:graph 0 x31 :0 the limit and (b) find the limit if it exists. x: -2 x: -2 x + 2 x + 2 In Exercises 27–30, use the given graph to find 2 the limits or to explain 2 x + 7x + 12 x 9 lim ƒ1x2 lim ƒ1x2 (b) , and (c) if they exist. ƒ1x2(a) = 11 lim + x2+1/xƒ 33. 31. ƒ1x2 (a) lim sin9lim x x:0+ 2 27. 11. 2y12. why the limits do notxexist. x:0 lim 32. x: -1 (g) + 7x + 12 x x : 2 2 2 1,ƒ1x2 In Exercises 19–22, the-fact that xlim along with the x: -3 usex12. x:3= 1/x 2 9 x + 2x 15 35. Group Activit ƒ1x2 = 11 = 11 + x2 33. 34. lim lim 11. lim ƒ1x2 27. (a) lim x (b) lim+ ƒ1x2 y 23 2 x:0 lim ƒ1 (b) 17. 18.x:3lim x:--31x 3 2x - 15 3 x:2 x:2 lim g1x2 = 4. Fin x 9 x + 3 2 2 x:0 x:0 x:0 ƒ1 lim x + 1 x 2x + x 2 35. Group Activity Assume that x:4 (i) x limit properties, to find the following limits. (b) lim+ ƒ1x23 2 ƒ1x2 3 (c) lim x:4 lim x x:2 = 4. Find the (a) limit. 2 lim 1g1x2 +ƒ1 x + 113. x : x 3 -1 14. 2x 2 x:2 +limx - x:4 2lim xg1x2 lim (c) x: -1 x + 1 2 x : 4 x:0 lim lim 13. 14. 2 (c) lim ƒ1x2 sin x sin x sin 3x (a) 32. (a)xƒ x:2 2 limthe (b)2 lim x: -1 x:2 20.=xlim x + lim 1use the fact -, 2along 2 3 ƒ xxwith 21 that lim 1 In Exercises 19–22, 19. -x:4 4 ƒ 1g1x2 + 22 (c) (d) x 4 x:4 ƒ1 1 0 lim g 1x2 lim 2 x:0 x:0 limx x:0 x 2 16. 15.2xlim- x x : 4 x:0 2 SECTIONSECTION 10.3 EXERCISES 10.3 EXERCISES ƒ x ƒ SECTION 10.3 EXERCISES x (b) lim+ ƒ1x2 x x:1 7. lim 1e sin x2 8. lim ln asin b x:0 x:p 2 (c) lim ƒ1x2 mbers with a gray background indicate problems that 29. (a) lim- ƒ1x2 y x:1 2 x : 3 x - 1 have designed to be solved without a calculator. 2 Example 2 Explain why-the be found 22 limit cannot10. lim by direct 9. lim 1x lim+ ƒ1x2 and find algebraically (b) substitution x:a x:a x 2 + 1 4 x :3 1–10, find the limit by direct if thesubstitution limit exists.if it exists. In Exercises 31 and3 32, the graph of a functio ƒ1x2 (c) lim 2 In Exercises 11–18, 12 explain why you cannot use substitution to find x:3 x 1x - 122 - 12(a) 2. lim 1x Which of the statements about the function ar 1 x : 3 the limit and (b) find the limit algebraically if it exists. x SECTION a. 123 41 0ƒ1x2 3 3 lim = 31. (a) 1x - 2x + 32 4. lim 1x 2 - x + 52 x: -1 + x2 - 9 x : -2 x + 7x + 12 lim 11. lim 12. y = 0 (b) lim- ƒ1x2 ƒ1x2 x:3 x 2 +30. lim-3 1x -x 242-2/39 1x + 5 6. x: 2x (a) - 15xlim x:0 : 1x : -2 x 3 + 1x x 3 - 2x 2(b) + x lim - 2 ƒ1x2 = 1 (c) lim- ƒ1x2 4 x x:0 lim lim 13. 14. x : 1+ 1e sin x2 ln a sin b 8. lim x:2 x + 763 12 x - 2 78.qxd 1/20/10 3:35 PM xx: : p-1Page 2 = lim+ ƒ1x2 (d) lim- ƒ1x2 (c) lim ƒ1x2 2 1 x:0 x:0 2 x : 1 x 4 ƒ ƒ x 2x - 4 x 1 lim lim 15. 16. lim(a) ƒ1x2 exists (e)29. (f) 1 2 ƒ1x2 3 0lim Exercise numbers with a gray background indicate problems that 1x 2 - 22 10. lim x: -2 x + 2 x: -2 x + 2 x:0 x x:3 2 x : a xhave the authors + 1designed to be solved without a calculator. lim(b) ƒ1x2 =ƒ1x2 1ƒ1x2 y = 31 and 32, the graph of a(g) function is given. (h) x lim xIn-Exercises 2 x:0 11–18, (a) explain why you cannot use toby find x:3 + lim substitution lim 1x - find 3substitution 17. 18. In Exercises 1–10, the limit direct if it exists. true=and x:0 x:0 Which ƒ1x2 0 which are false? (i) limare (j) x 2 of the statements about the function (b) find the limit algebraically if it exists. 2 x:1(c) lim ƒ1x2 x 12 y x:3 lim lim x 1x 12 1x 12 1. 2. lim + ƒ1x2SECTION = 1 32.10.3 sinx:3 x 31. (a) x : 2 2 lim ƒ1x2 = 1 (a) More on Limits 763 x: -1 x + 7x + 12 x = 1, along with-1the In Exercises 19–22, use9the fact that lim x: -1 + 12. lim 2 3 x:0 x 3 (b) lim ƒ1x2 = 0 2 lim 1x 2x + 32 lim 1x x + 52 3. 4. 3 x : 3 3 exist b. x:2x +to 2x x - 9 (b) lim ƒ1x2 does not x : 0limit properties, find-the15following limits.x: -2 x:2 3 3 2 x + 1 - 2x + x - 2 lim ƒ1x2 = 1 (c) 2/3 lim lim x1x 5. lim 6. lim lim(a) ƒ1x2 = 2ƒ1x2 (c)30. x sin x+ 5 sin-23x1x - 42x : 0 14. x:1 - –1 1 2 3 x:2 x: x:2 lim 19. 20. 1 x + 1 xlim :2 x 2 SECTION 10.3x:0EXERCISES (d)x lim- ƒ1x2 = lim+ ƒ1x2 (d) lim x:0 x lim= ƒ1x2 (b)ƒ1x2 2x 2x2 - x 2 x:0 2 x:1 - x:1 + - 4x2ƒ ƒ x2 sin 7. lim 1e 8. lim ln asin xb: 0 x - 4 x x + sin x(e) 2 lim ƒ1x2 exists x:p lim sin 16. x:0 lim ƒ1x2 1= 0 lim(c) (e) (f) umbers problems that 29. (a) lim 21. xlim 22. lim yƒ1x2 2 x + 2with a gray background : -2 indicate x : 0 ƒ1x2 0 lim=ƒ1x2 x + 2 x:1x+: x:1 x:3 x:0 x:0 x 2x rs have designed to be solved without a calculator. lim ƒ1x2 = 1 (g) x 2 -(b) 1 xlim lim lim ƒ1x2ƒ1x2 (f) (h) does=not1 exist 22 x : 0+ ƒ1x2 x:1x4: 1 lim 1x 22 lim 9. 10. In Exercises 23–26, find the limits. x:3 lim 1x 3 18. es 1–10, find the limit by direct substitution if it exists. x:a x 2 + (i) xx:a :0 1 lim ƒ1x2 = 0 ƒ1x2 = 2 ƒ1x2 xx2 lim3lim ƒ1x2 = lim (g) (j) lim x :x1 ƒ1x2 e 1x 3 sin x -(c)4 cos 2 12 In Exercises 32,- the graph o x:0x2+: 2 31 and x:0 x:3 lim m x 1x - 12 1x 12 2. lim lim cannot use substitution to find 23. x:3 24. you In Exercises 11–18, (a) explain why sin x : -1 lim ƒ1x2 = 1 32. (a) x every x:0 5 sin x + cos x + 22 with the ƒ1x2statements (h) lim exists for in 1Which of the aboutc the f + 1, along 19–22, use the fact that lim x:0 log=431x x:c0 1 2 3 4 find the limit algebraically if it exists.x : -1 x4. : 0 and x (b) m 1x 3 - 2x + 32 the limit lim 1x x + 52 1x + 9(b) lim ƒ1x2 does not exist (i)31. :2 x: -2 ln12x2 lim(a) ƒ1x2 lim exists ƒ1x2 for every = 1c in 11 2 ies, to find the following25. limits. lim x 2 2+ 7x2/3+ 12 26. lim x9: 2 x:c yx: -1 + y x(a) - lim 30. ƒ1x2 lim m 1x + 5 1x 42 6. x:p/2 x:27 log x sin x lim 11. x: 12. lim 23 (c) x:1 :2 sin x -2 lim- ƒ1x2 = 2 In Exercises 33 (b) and 34, a graph sin 3xx 2 - 9 limuseƒ1x2 = 0of ƒ to fi x: -3 x:3 x + 2x x-:15 c. lim 27–30, use x:0 In 20. Exercises to2+explain (b) orlim ƒ1x2 x the given graph to find the limits 4 2x x : 0 lim ƒ1x2 lim ƒ1x2 (b) , and (c) if they exist x x:1 ƒ1x2 = 2 3 lim lndo m2x1e -sinxx2 3 not exist. why 8. the x:p limits x:0 + x:0ƒ1x2 x asin + 12 b x 3 - (d) 2x 2xlim + 1x- - 2 = 1 (c)2 lim 2 :0 : (c) lim ƒ1x2 x:0 -1/x 13. limx + sin x 14. lim sin2 x 1 33. ƒ1x2 = 11 + x2 34. ƒ1x2 x: -1lim x:2 (e) y x- + 1 x x:1 -lim 2 ƒ1x2 = 1 ƒ1x2 27. lim (a) 22. x1 2=3 limx ƒ1x –1 ƒ1x2 (d)0 1lim x : 1+ x 2 2x - 1 2 3 x : 0 x:2 x 2 x:0 x:0 +that m 1x - 22 10. lim 2x 2 - 4 35. Group Activity Assume x 2 (f - )4 ƒlim ƒ1x2 does not exist ƒ (b) :a x:a xlim++ƒ1x2 x:1 23–26, find the limits. 15. limx:2 1 16. 3lim lim g1x2 4. Find limit. lim ƒ1x2 exists (e) y = =x:0 ƒ1x2 In Exercises 31 and 32, the graph of a function is the given. x:4 x: -2 x + 2 2 -2 x +(g)2 lim+ ƒ1x2 = lim- ƒ1x2 es 11–18, (a) explain why you(c) cannot lim use ƒ1x2substitution to find x: x e - 1x 3 sin x - 4 cos x Which of the xstatements about true and+ƒ1x2 which x:2 :0 x :the 0 function lim(g) 1g1x2 22 =are1 false? (a) are (b) lim 24. lim if it exists. nd (b) find the limit algebraically 1 x - 2 x:4 x x:0 x : 0 log41x + 22 5 sin x + cos x lim 1 1, 12 ƒ1x2 (h) exists for every c in . y 17. lim 1x 2- 3 18. lim 31.2 (a) xlim ƒ1x2 = 1 2 x + : c x:0 x:0 x + 7x + 12 x - 9 lim ƒ1x2 = 0 (i)g21x2 1 x 2 3 x: -1 0 lim (c) (d) + 9 m ln12x22 12. lim 1x x:1 lim ƒ1x2 11, 32 (i) exists for every c in . x:4 x 2 lim ƒ1x2 = 0 (b) : lim x + 2x - 15 m -3 2 x - 9 26. x:3 3 x : c sin x x:0 p/2 sin x x : 27 log3 x lim ƒ1x2 = 1 32. (a) + = (c) 1, along In Exercises 19–22, use with 3 ƒ1x2 2 the fact that lim In 28. (a) xlim 36.ofGroup Assume that x: -1 y Exercises ƒ1x2 lim 33 and 34,the ƒ to findActivity (a) , x3 + 1 lim ƒ1x2 =use 1 a graph x:3 -- 2x + x - 2 x x : 0x:0 27–30, use the given graph find the limits or to explain x:0 x m 14. tolim lim g1x2 = -3 . Find the limit. 1 2 3 –1 lim ƒ1x2 (b) does not exis (b) lim+ ƒ1x2, and (c) lim ƒ1x2 if they exist. : -1 x + 1 x:2 lim to xfind - 2the following limits. ƒ1x2 (b) x:a limit properties, x:2 ts do not exist. x : 0(d) lim- ƒ1x2 = x : lim 0 + ƒ1x2 x:3 + 2 x:0 x:0 3 lim(c) 1ƒ1x2 g1x22 (a) ƒ1x2 (b) -x 4 ƒ ƒ xsin x 2ƒ1x2 - 4 = 2 = lim 11 ++ƒ1x2 x21/12x2 33.3xƒ1x2 = 11 + x21/x 34. ƒ1x2 (c) lim 2 sin x:a x y lim m lim 16. lim lim ƒ1x2 exists ƒ1x2 = 0 (e) (f) x:2 x:3 1 lim-2 x2 + 2 19. 20. lim : 2x + 2 :x-2 x: d. x:0 x:0 x:0 x x:0 35. x Group ƒ1x2 ƒ1x2 = - 1=and lim lim Activity Assume that(d) 2x - x 2 (d) 0 1 2 3 4 (g) lim ƒ1x2 lim (c) -+ 112 x13g1x2 :ƒ1x2 4x:1= lim lim ƒ1x2 = 1 (h) 3 x -2 2 x : 2+ limx:0 g1x2 = 4. Find the limit. x:a x:1 x + sin x : 4x mlim 1xƒ1x2 - 3 18. lim2 sin2 x lim+=ƒ1x2 (e) ƒ1x2 lim x lim 21. 22. :0 x:0 lim = 0+ 22 2 = 1 (i) (j)(b)lim - x:1 x:2 limƒ1x2 1g1x2 lim xƒ1x2 (a)x:1 x:0 x:0 x 2x x:2 1 x:4 x:4 sin x lim ƒ1x2 does not exis (f) 32. (a) lim + ƒ1x2 = 1 lim = 1, along es 19–22, use the factIn that with the xlimits. x:1g1x2 Exercises 23–26, find the x: -1 2 1 2 3 x:0 x0 (c) lim g 1x2 (d) lim lim ƒ1x2 (g) lim x : 4ƒ1x2 does not exist x : 4 ƒ1x2 - 1 = lim- ƒ1x 3 sin(b) x -x:2 4 cos x rties, to find the following limits. ex - 1x x:0 + x:0 lim lim 23. 24. y lim- ƒ1x2 ƒ1x2 = 2 lim 36. Group Activity Assume that and y x:0 x:0 lim ƒ1x2 = 2 (c) log 1x + 22 5 sin x + cos x lim ƒ1x2 (h) exists for eve sin 3x 4 x : 3sin x x : ax:c m 2 20. lim limx:2 g1x2 = - 3. Find the limit. :0lim x:0 ln12x2 x 3 x+: 9 a lim ƒ1x2 = 2 2x ƒ1x2 - x 1x(d) (i) lim ƒ1x2 x : 3+ 2 # exists for eve x:1 lim 25. lim 26. x:c 2 3 x + sin lim 1ƒ1x2 + g1x22 lim 1ƒ1x2 g1x22 (a) (b) 2 x x:p/2 x:27 sin x log(e) ƒ1x2 3 x lim x : a ƒ1x2 = 1 x:a x mxlim 22. lim21 sin x + :3 In Exercises 33 and 34, 1 2 3use a graph –1 x:1 :0 x:0 x 2x ƒ1x2 x In Exercises027–30, limits or to does explain 1 2 3 4use the given graph to find the(f) 13g1x2 +not 12exist (b) lim lim , and (c) lim ƒ1x2 if limƒ1x2 (c) lim (d)+ ƒ1x2 SECTION 10.3 EXERCISES