Solution Manual for Calculus, 10th Edition

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Limits and ContinuityExercise Set 1.11. (a)3(b)3(c)3(d)32. (a)0(b)0(c)0(d)03. (a)−1(b)3(c)does not exist(d)14. (a)2(b)0(c)does not exist(d)25. (a)0(b)0(c)0(d)36. (a)1(b)1(c)1(d)07. (a)−∞(b)−∞(c)−∞(d)18. (a)+∞(b)+∞(c)+∞(d)can not be found from graph9. (a)+∞(b)+∞(c)2(d)2(e)−∞(f )x=−2,x= 0,x= 210. (a)does not exist(b)−∞(c)0(d)−1(e)+∞(f )3(g)x=−2,x= 211. (i)−0.01−0.001−0.00010.00010.0010.010.99501660.99950020.99995001.00005001.00050021.0050167(ii)0.9951.005-0.010.01The limit appears to be 1.12. (i)−0.1−0.01−0.0010.0010.010.12.01357922.00013342.00000132.00000132.00013342.0135792(ii)2.01421.01.0-The limit appears to be 2.39

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40Chapter 113. (a)21.51.11.011.00100.50.90.990.9990.14290.21050.30210.33000.33301.00000.57140.36900.33670.33371002The limit is 1/3.(b)21.51.11.011.0011.00010.42861.05266.34466.33666.36666.350012The limit is +∞.(c)00.50.90.990.9990.9999−1−1.7143−7.0111−67.001−667.0−6667.00-5001The limit is−∞.14. (a)−0.25−0.1−0.001−0.00010.00010.0010.10.250.53590.51320.50010.50000.50000.49990.48810.47210.60-0.250.25The limit is 1/2.(b)0.250.10.0010.00018.472120.4882000.520001

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Exercise Set 1.141100000.25The limit is +∞.(c)−0.25−0.1−0.001−0.0001−7.4641−19.487−1999.5−200000-100-0.250The limit is−∞.15. (a)−0.25−0.1−0.001−0.00010.00010.0010.10.252.72662.95523.00003.00003.00003.00002.95522.726632-0.250.25The limit is 3.(b)0−0.5−0.9−0.99−0.999−1.5−1.1−1.01−1.00111.75526.216154.87541.1−0.1415−4.536−53.19−539.560-60-1.50The limit does not exist.16. (a)0−0.5−0.9−0.99−0.999−1.5−1.1−1.01−1.0011.55741.09261.00331.00001.00001.09261.00331.00001.00001.51-1.50The limit is 1.

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42Chapter 1(b)−0.25−0.1−0.001−0.00010.00010.0010.10.251.97942.41322.50002.50002.50002.50002.41321.97942.52-0.250.25The limit is 5/2.17.False; definef(x) =xforx6=aandf(a) =a+ 1. Then limx→af(x) =a6=f(a) =a+ 1.18.True; by 1.1.3.19.False; definef(x) = 0 forx <0 andf(x) =x+ 1 forx≥0. Then the left and right limits exist but are unequal.20.False; definef(x) = 1/xforx >0 andf(0) = 2.27.msec=x2−1x+ 1=x−1 which gets close to−2 asxgets close to−1, thusy−1 =−2(x+ 1) ory=−2x−1.28.msec=x2x=xwhich gets close to 0 asxgets close to 0, thusy= 0.29.msec=x4−1x−1=x3+x2+x+ 1 which gets close to 4 asxgets close to 1, thusy−1 = 4(x−1) ory= 4x−3.30.msec=x4−1x+ 1=x3−x2+x−1 which gets close to−4 asxgets close to−1, thusy−1 =−4(x+1) ory=−4x−3.31. (a)The length of the rod while at rest.(b)The limit is zero. The length of the rod approaches zero as its speed approachesc.32. (a)The mass of the object while at rest.(b)The limiting mass as the velocity approaches the speed of light; the mass is unbounded.33. (a)3.52.5–11The limit appears to be 3.(b)3.52.5–0.0010.001The limit appears to be 3.

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Exercise Set 1.243(c)3.52.5–0.0000010.000001The limit does not exist.Exercise Set 1.21. (a)By Theorem 1.2.2, this limit is 2 + 2·(−4) =−6.(b)By Theorem 1.2.2, this limit is 0−3·(−4) + 1 = 13.(c)By Theorem 1.2.2, this limit is 2·(−4) =−8.(d)By Theorem 1.2.2, this limit is (−4)2= 16.(e)By Theorem 1.2.2, this limit is3√6 + 2 = 2.(f )By Theorem 1.2.2, this limit is2(−4) =−12 .2. (a)By Theorem 1.2.2, this limit is 0 + 0 = 0.(b)The limit doesn’t exist because limfdoesn’t exist and limgdoes.(c)By Theorem 1.2.2, this limit is−2 + 2 = 0.(d)By Theorem 1.2.2, this limit is 1 + 2 = 3.(e)By Theorem 1.2.2, this limit is 0/(1 + 0) = 0.(f )The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.(g)The limit doesn’t exist because√f(x) is not defined for 0< x <2.(h)By Theorem 1.2.2, this limit is√1 = 1.3.By Theorem 1.2.3, this limit is 2·1·3 = 6.4.By Theorem 1.2.3, this limit is 33−3·32+ 9·3 = 27.5.By Theorem 1.2.4, this limit is (32−2·3)/(3 + 1) = 3/4.6.By Theorem 1.2.4, this limit is (6·0−9)/(03−12·0 + 3) =−3.7.After simplification,x4−1x−1=x3+x2+x+ 1, and the limit is 13+ 12+ 1 + 1 = 4.8.After simplification,t3+ 8t+ 2=t2−2t+ 4, and the limit is (−2)2−2·(−2) + 4 = 12.9.After simplification,x2+ 6x+ 5x2−3x−4 =x+ 5x−4 , and the limit is (−1 + 5)/(−1−4) =−4/5.

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44Chapter 110.After simplification,x2−4x+ 4x2+x−6=x−2x+ 3 , and the limit is (2−2)/(2 + 3) = 0.11.After simplification, 2x2+x−1x+ 1= 2x−1, and the limit is 2·(−1)−1 =−3.12.After simplification, 3x2−x−22x2+x−3 = 3x+ 22x+ 3 , and the limit is (3·1 + 2)/(2·1 + 3) = 1.13.After simplification,t3+ 3t2−12t+ 4t3−4t=t2+ 5t−2t2+ 2t, and the limit is (22+ 5·2−2)/(22+ 2·2) = 3/2.14.After simplification,t3+t2−5t+ 3t3−3t+ 2=t+ 3t+ 2 , and the limit is (1 + 3)/(1 + 2) = 4/3.15.The limit is +∞.16.The limit is−∞.17.The limit does not exist.18.The limit is +∞.19.The limit is−∞.20.The limit does not exist.21.The limit is +∞.22.The limit is−∞.23.The limit does not exist.24.The limit is−∞.25.The limit is +∞.26.The limit does not exist.27.The limit is +∞.28.The limit is +∞.29.After simplification,x−9√x−3 =√x+ 3, and the limit is√9 + 3 = 6.30.After simplification,4−y2− √y= 2 +√y, and the limit is 2 +√4 = 4.31. (a)2(b)2(c)232. (a)does not exist(b)1(c)433.True, by Theorem 1.2.2.34.False; e.g. limx→0x2x= 0.

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Exercise Set 1.24535.False; e.g.f(x) = 2x, g(x) =x, so limx→0f(x) = limx→0g(x) = 0, but limx→0f(x)/g(x) = 2.36.True, by Theorem 1.2.4.37.After simplification,√x+ 4−2x=1√x+ 4 + 2 , and the limit is 1/4.38.After simplification,√x2+ 4−2x=x√x2+ 4 + 2 , and the limit is 0.39. (a)After simplification,x3−1x−1=x2+x+ 1, and the limit is 3.(b)yx4140. (a)After simplification,x2−9x+ 3=x−3, and the limit is−6, so we need thatk=−6.(b)On its domain (all real numbers),f(x) =x−3.41. (a)Theorem 1.2.2 doesn’t apply; moreover one cannot subtract infinities.(b)limx→0+(1x−1x2)=limx→0+(x−1x2)=−∞.42. (a)Theorem 1.2.2 assumes thatL1andL2are real numbers, not infinities. It is in general not true that ”∞·0 = 0 ”.(b)1x−2x2+ 2x=x2x(x2+ 2x) =1x+ 2 forx6= 0, so that limx→0(1x−2x2+ 2x)= 12 .43.Forx6= 1,1x−1−ax2−1 =x+ 1−ax2−1and for this to have a limit it is necessary that limx→1(x+ 1−a) = 0, i.e.a= 2. For this value,1x−1−2x2−1 =x+ 1−2x2−1=x−1x2−1 =1x+ 1 and limx→11x+ 1 = 12 .44. (a)For smallx, 1/x2is much bigger than±1/x.(b)1x+1x2=x+ 1x2. Since the numerator has limit 1 andx2tends to zero from the right, the limit is +∞.45.The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned realnumber. For example, letq(x) =x−x0and letp(x) =a(x−x0)nwherentakes on the values 0,1,2.46.If on the contrary limx→ag(x) did exist then by Theorem 1.2.2 so would limx→a[f(x) +g(x)], and that would be acontradiction.47.Clearly,g(x) = [f(x) +g(x)]−f(x). By Theorem 1.2.2, limx→a[f(x) +g(x)]−limx→af(x) = limx→a[f(x) +g(x)−f(x)] =limx→ag(x).

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46Chapter 148.By Theorem 1.2.2, limx→af(x) =(limx→af(x)g(x))limx→ag(x) =(limx→af(x)g(x))·0 = 0, since limx→af(x)g(x) exists.Exercise Set 1.31. (a)−∞(b)+∞2. (a)2(b)03. (a)0(b)−14. (a)does not exist(b)05. (a)3 + 3·(−5) =−12(b)0−4·(−5) + 1 = 21(c)3·(−5) =−15(d)(−5)2= 25(e)3√5 + 3 = 2(f )3/(−5) =−3/5(g)0(h)The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.6. (a)2·7−(−6) = 20(b)6·7 + 7·(−6) = 0(c)+∞(d)−∞(e)3√−42(f )−6/7(g)7(h)−7/127. (a)x0.10.010.0010.00010.000010.000001f(x)1.4711281.5607971.5697961.5706961.5707861.570795The limit appears to be≈1.57079. . ..(b)The limit isπ/2.8.x101001000100001000001000000f(x)1.2589251.0471291.0069321.0009211.0001151.000014The limit appears to be 1.9.The limit is−∞, by the highest degree term.10.The limit is +∞, by the highest degree term.11.The limit is +∞.12.The limit is +∞.13.The limit is 3/2, by the highest degree terms.14.The limit is 5/2, by the highest degree terms.15.The limit is 0, by the highest degree terms.16.The limit is 0, by the highest degree terms.17.The limit is 0, by the highest degree terms.18.The limit is 5/3, by the highest degree terms.19.The limit is−∞, by the highest degree terms.20.The limit is +∞, by the highest degree terms.21.The limit is−1/7, by the highest degree terms.22.The limit is 4/7, by the highest degree terms.

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Exercise Set 1.34723.The limit is3√−5/8 =−3√5/2, by the highest degree terms.24.The limit is3√3/2 , by the highest degree terms.25.√5x2−2x+ 3=√5−2x2−1−3xwhenx <0. The limit is−√5 .26.√5x2−2x+ 3=√5−2x21 +3xwhenx >0. The limit is√5 .27.2−y√7 + 6y2=−2y+ 1√7y2+ 6wheny <0. The limit is 1/√6 .28.2−y√7 + 6y2=2y−1√7y2+ 6wheny >0. The limit is−1/√6 .29.√3x4+xx2−8=√3 +1x31−8x2whenx <0. The limit is√3 .30.√3x4+xx2−8=√3 +1x31−8x2whenx >0. The limit is√3 .31.limx→+∞(√x2+ 3−x)√x2+ 3 +x√x2+ 3 +x=limx→+∞3√x2+ 3 +x= 0, by the highest degree terms.32.limx→+∞(√x2−3x−x)√x2−3x+x√x2−3x+x=limx→+∞−3x√x2−3x+x=−3/2, by the highest degree terms.33.limx→−∞1−ex1 +ex= 1−01 + 0 = 1.34.Divide the numerator and denominator byex:limx→+∞1−ex1 +ex=limx→+∞e−x−1e−x+ 1 = 0−10 + 1 =−1.35.Divide the numerator and denominator byex:limx→+∞1 +e−2x1−e−2x= 1 + 01−0 = 1.36.Divide the numerator and denominator bye−x:limx→−∞e2x+ 1e2x−1 = 0 + 10−1 =−1.37.The limit is−∞.38.The limit is +∞.39.x+ 1x= 1 + 1x, solimx→+∞(x+ 1)xxx=efrom Figure 1.3.4.40.(1 + 1x)−x=1(1 +1x)x, so the limit ise−1.41.False:limx→+∞(1 + 1x)2x=[limx→+∞(1 + 1x)x]2=e2.42.False;y= 0 is a horizontal asymptote for the curvey=exyetlimx→+∞exdoes not exist.43.True: for examplef(x) = sinx/xcrosses thex-axis infinitely many times atx=nπ, n= 1,2, . . ..

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48Chapter 144.False: if the asymptote isy= 0, thenlimx→±∞p(x)/q(x) = 0, and clearly the degree ofp(x) is strictly less than thedegree ofq(x). If the asymptote isy=L6= 0, thenlimx→±∞p(x)/q(x) =Land the degrees must be equal.45.It appears thatlimt→+∞n(t) = +∞, andlimt→+∞e(t) =c.46. (a)It is the initial temperature of the potato (400◦F).(b)It is the ambient temperature, i.e. the temperature of the room.47. (a)+∞(b)−548. (a)0(b)−649.limx→−∞p(x) = +∞. Whennis even,limx→+∞p(x) = +∞; whennis odd,limx→+∞p(x) =−∞.50. (a)p(x) =q(x) =x.(b)p(x) =x,q(x) =x2.(c)p(x) =x2,q(x) =x.(d)p(x) =x+ 3,q(x) =x.51. (a)No.(b)Yes, tanxand secxatx=nπ+π/2 and cotxand cscxatx=nπ, n= 0,±1,±2, . . ..52.Ifm > nthe limit is zero. Ifm=nthe limit iscm/dm. Ifn > mthe limit is +∞ifcndm>0 and−∞ifcndm<0.53. (a)Every value taken byex2is also taken byet:chooset=x2.Asxandtincrease without bound, so doeset=ex2. Thuslimx→+∞ex2=limt→+∞et= +∞.(b)Iff(t)→+∞(resp.f(t)→ −∞) thenf(t) can be made arbitrarily large (resp.small) by takingtlargeenough. But by considering the valuesg(x) whereg(x)> t, we see thatf(g(x)) has the limit +∞too (resp. limit−∞).Iff(t) has the limitLast→+∞the valuesf(t) can be made arbitrarily close toLby takingtlargeenough. But ifxis large enough theng(x)> tand hencef(g(x)) is also arbitrarily close toL.(c)Forlimx→−∞the same argument holds with the substitutiion ”xdecreases without bound” instead of ”xincreaseswithout bound”. Forlimx→c−substitute ”xclose enough toc, x < c”, etc.54. (a)Every value taken bye−x2is also taken byet: chooset=−x2. Asxincreases without bound andtdecreaseswithout bound, the quantityet=e−x2tends to 0. Thuslimx→+∞e−x2=limt→−∞et= 0.(b)Iff(t)→+∞(resp.f(t)→ −∞) thenf(t) can be made arbitrarily large (resp.small) by takingtsmallenough. But by considering the valuesg(x) whereg(x)< t, we see thatf(g(x)) has the limit +∞too (resp. limit−∞).Iff(t) has the limitLast→ −∞the valuesf(t) can be made arbitrarily close toLby takingtsmallenough. But ifxis large enough theng(x)< tand hencef(g(x)) is also arbitrarily close toL.(c)Forlimx→−∞the same argument holds with the substitutiion ”xdecreases without bound” instead of ”xincreaseswithout bound”. Forlimx→c−substitute ”xclose enough toc, x < c”, etc.55.t= 1/x,limt→+∞f(t) = +∞.56.t= 1/x,limt→−∞f(t) = 0.57.t= cscx,limt→+∞f(t) = +∞.58.t= cscx,limt→−∞f(t) = 0.59.Lett= lnx. Thentalso tends to +∞, and ln 2xln 3x=t+ ln 2t+ ln 3 , so the limit is 1.60.Witht=x−1,[ln(x2−1)−ln(x+ 1)] = ln(x+ 1) + ln(x−1)−ln(x+ 1) = lnt, so the limit is +∞.

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Exercise Set 1.34961.Sett=−x, then getlimt→−∞(1 + 1t)t=eby Figure 1.3.4.62.Witht=x/2,limx→+∞(1 + 2x)x=(limt→+∞[1 + 1/t]t)2=e263.From the hint,limx→+∞bx=limx→+∞e(lnb)x=0ifb <1,1ifb= 1,+∞ifb >1.64.It suffices by Theorem 1.1.3 to show that the left and right limits at zero are equal toe.(a)limx→+∞(1 +x)1/x=limt→0+(1 + 1/t)t=e.(b)limx→−∞(1 +x)1/x=limt→0−(1 + 1/t)t=e.65. (a)481216204080120160200tv(b)limt→∞v= 190(1−limt→∞e−0.168t)= 190, so the asymptote isv=c= 190 ft/sec.(c)Due to air resistance (and other factors) this is the maximum speed that a sky diver can attain.66. (a)p(1990) = 525/(1 + 1.1) = 250 (million).(b)192020002080250500tP(c)limt→∞p(t) =5251 + 1.1 limt→∞e−0.02225(t−1990)= 525 (million).(d)The population becomes stable at this number.67. (a)n2345671 + 10−n1.011.0011.00011.000011.0000011.00000011 + 10n101100110001100001100000110000001(1 + 10−n)1+10n2.73192.71962.71842.71832.718282.718282The limit appears to bee.(b)This is evident from the lower left term in the chart in part (a).(c)The exponents are being multiplied bya, so the result isea.

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50Chapter 168. (a)f(−x) =(1−1x)−x=(x−1x)−x=(xx−1)x,f(x−1) =(xx−1)x−1=(x−1x)f(−x).(b)limx→−∞(1 + 1x)x=limx→+∞f(−x) =[limx→+∞xx−1]limx→+∞f(x−1) =limx→+∞f(x−1) =e.69.After a long division,f(x) =x+ 2 +2x−2 , solimx→±∞(f(x)−(x+ 2)) = 0 andf(x) is asymptotic toy=x+ 2.The only vertical asymptote is atx= 2.–12–63915–15–9–33915xx= 2yy=x+ 270.After a simplification,f(x) =x2−1 + 3x, solimx→±∞(f(x)−(x2−1)) = 0 andf(x) is asymptotic toy=x2−1.The only vertical asymptote is atx= 0.–4–224–2135xyy=x2– 171.After a long division,f(x) =−x2+1+2x−3 , solimx→±∞(f(x)−(−x2+1)) = 0 andf(x) is asymptotic toy=−x2+1.The only vertical asymptote is atx= 3.–4–224–12–6612xx= 3yy= –x2+ 172.After a long division,f(x) =x3+32(x−1)−32(x+ 1) , solimx→±∞(f(x)−x3) = 0 andf(x) is asymptotic toy=x3.The vertical asymptotes are atx=±1.

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Exercise Set 1.451–22–15515xyy=x3x= 1x= –173.limx→±∞(f(x)−sinx) = 0 sof(x) is asymptotic toy= sinx. The only vertical asymptote is atx= 1.–428–435xyy= sinxx= 1Exercise Set 1.41. (a)|f(x)−f(0)|=|x+ 2−2|=|x|<0.1 if and only if|x|<0.1.(b)|f(x)−f(3)|=|(4x−5)−7|= 4|x−3|<0.1 if and only if|x−3|<(0.1)/4 = 0.025.(c)|f(x)−f(4)|=|x2−16|< if|x−4|< δ.We getf(x) = 16 += 16.001 atx= 4.000124998, whichcorresponds toδ= 0.000124998; andf(x) = 16−= 15.999 atx= 3.999874998, for whichδ= 0.000125002. Usethe smallerδ: thus|f(x)−16|< provided|x−4|<0.000125 (to six decimals).2. (a)|f(x)−f(0)|=|2x+ 3−3|= 2|x|<0.1 if and only if|x|<0.05.(b)|f(x)−f(0)|=|2x+ 3−3|= 2|x|<0.01 if and only if|x|<0.005.(c)|f(x)−f(0)|=|2x+ 3−3|= 2|x|<0.0012 if and only if|x|<0.0006.3. (a)x0= (1.95)2= 3.8025, x1= (2.05)2= 4.2025.(b)δ= min (|4−3.8025|,|4−4.2025|) = 0.1975.4. (a)x0= 1/(1.1) = 0.909090. . . , x1= 1/(0.9) = 1.111111. . .(b)δ= min(|1−0.909090|,|1−1.111111|) = 0.0909090. . .5.|(x3−4x+5)−2|<0.05 is equivalent to−0.05<(x3−4x+5)−2<0.05, which means 1.95< x3−4x+5<2.05. Nowx3−4x+5 = 1.95 atx= 1.0616, andx3−4x+5 = 2.05 atx= 0.9558. Soδ= min (1.0616−1,1−0.9558) = 0.0442.

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52Chapter 12.21.90.91.16.√5x+ 1 = 3.5 atx= 2.25,√5x+ 1 = 4.5 atx= 3.85, soδ= min(3−2.25,3.85−3) = 0.75.50247.With the TRACE feature of a calculator we discover that (to five decimal places) (0.87000,1.80274) and (1.13000,2.19301)belong to the graph. Setx0= 0.87 andx1= 1.13. Since the graph off(x) rises from left to right, we see that ifx0< x < x1then 1.80274< f(x)<2.19301, and therefore 1.8< f(x)<2.2. So we can takeδ= 0.13.8.From a calculator plot we conjecture thatlimx→0f(x) = 2.Using the TRACE feature we see that the points(±0.2,1.94709) belong to the graph. Thus if−0.2< x <0.2, then 1.95< f(x)≤2 and hence|f(x)−L|<0.05<0.1 =.9.|2x−8|= 2|x−4|<0.1 when|x−4|<0.1/2 = 0.05 =δ.10.|(5x−2)−13|= 5|x−3|<0.01 when|x−3|<0.01/5 = 0.002 =δ.11.Ifx6= 3, then∣∣∣∣x2−9x−3−6∣∣∣∣=∣∣∣∣x2−9−6x+ 18x−3∣∣∣∣=∣∣∣∣x2−6x+ 9x−3∣∣∣∣=|x−3|<0.05 when|x−3|<0.05 =δ.12.Ifx6=−1/2, then∣∣∣∣4x2−12x+ 1−(−2)∣∣∣∣=∣∣∣∣4x2−1 + 4x+ 22x+ 1∣∣∣∣=∣∣∣∣4x2+ 4x+ 12x+ 1∣∣∣∣=|2x+ 1|= 2|x−(−1/2)|<0.05 when|x−(−1/2)|<0.025 =δ.13.Assumeδ≤1. Then−1< x−2<1 means 1< x <3 and then|x3−8|=|(x−2)(x2+ 2x+ 4)|<19|x−2|, sowe can chooseδ= 0.001/19.14.Assumeδ≤1. Then−1< x−4<1 means 3< x <5 and then|√x−2|=∣∣∣∣x−4√x+ 2∣∣∣∣<|x−4|√3 + 2 , so we can chooseδ= 0.001·(√3 + 2).15.Assumeδ≤1. Then−1< x−5<1 means 4< x <6 and then∣∣∣∣1x−15∣∣∣∣=∣∣∣∣x−55x∣∣∣∣<|x−5|20, so we can chooseδ= 0.05·20 = 1.16.||x| −0|=|x|<0.05 when|x−0|<0.05 =δ.17.Let >0 be given. Then|f(x)−3|=|3−3|= 0< regardless ofx, and hence anyδ >0 will work.18.Let >0 be given. Then|(x+ 2)−6|=|x−4|< providedδ=(although any smallerδwould work).

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Exercise Set 1.45319.|3x−15|= 3|x−5|< if|x−5|< /3,δ=/3.20.|7x+ 5 + 2|= 7|x+ 1|< if|x+ 1|< /7,δ=/7.21.∣∣∣∣2x2+xx−1∣∣∣∣=|2x|< if|x|< /2,δ=/2.22.∣∣∣∣x2−9x+ 3−(−6)∣∣∣∣=|x+ 3|< if|x+ 3|< ,δ=.23.|f(x)−3|=|x+ 2−3|=|x−1|< if 0<|x−1|< ,δ=.24.|9−2x−5|= 2|x−2|< if 0<|x−2|< /2,δ=/2.25.If >0 is given, then takeδ=; if|x−0|=|x|< δ, then|x−0|=|x|< .26.Ifx <2 then|f(x)−5|=|9−2x−5|= 2|x−2|< if|x−2|< /2, δ1=/2. Ifx >2 then|f(x)−5|=|3x−1−5|=3|x−2|< if|x−2|< /3, δ2=/3 Now letδ= min(δ1, δ2) then for anyxwith|x−2|< δ,|f(x)−5|< .27.For the first part, let >0. Then there existsδ >0 such that ifa < x < a+δthen|f(x)−L|< . For the leftlimit replacea < x < a+δwitha−δ < x < a.28. (a)Given >0 there existsδ >0 such that if 0<|x−a|< δthen||f(x)−L| −0|< , or|f(x)−L|< .(b)From part (a) it follows that|f(x)−L|< is the defining condition for each of the two limits, so the twolimit statements are equivalent.29. (a)|(3x2+ 2x−20−300|=|3x2+ 2x−320|=|(3x+ 32)(x−10)|=|3x+ 32| · |x−10|.(b)If|x−10|<1 then|3x+ 32|<65, since clearlyx <11.(c)δ= min(1, /65);|3x+ 32| · |x−10|<65· |x−10|<65·/65 =.30. (a)∣∣∣∣283x+ 1−4∣∣∣∣=∣∣∣∣28−12x−43x+ 1∣∣∣∣=∣∣∣∣−12x+ 243x+ 1∣∣∣∣=∣∣∣∣123x+ 1∣∣∣∣· |x−2|.(b)If|x−2|<4 then−2< x <6, soxcan be very close to−1/3, hence∣∣∣∣123x+ 1∣∣∣∣is not bounded.(c)If|x−2|<1 then 1< x <3 and 3x+ 1>4, so∣∣∣∣123x+ 1∣∣∣∣<124= 3.(d)δ= min(1, /3);∣∣∣∣123x+ 1∣∣∣∣· |x−2|<3· |x−2|<3·/3 =.31.Ifδ <1 then|2x2−2|= 2|x−1||x+ 1|<6|x−1|< if|x−1|< /6, soδ= min(1, /6).32.Ifδ <1 then|x2+x−12|=|x+ 4| · |x−3|<5|x−3|< if|x−3|< /5, soδ= min(1, /5).33.Ifδ <1/2 and|x−(−2)|< δthen−5/2< x <−3/2,x+ 1<−1/2,|x+ 1|>1/2; then∣∣∣∣1x+ 1−(−1)∣∣∣∣=|x+ 2||x+ 1|<2|x+ 2|< if|x+ 2|< /2, soδ= min(1/2, /2).34.Ifδ <1/4 and|x−(1/2)|< δthen∣∣∣∣2x+ 3x−8∣∣∣∣=|6x−3||x|<6|x−(1/2)|1/4= 24|x−(1/2)|< if|x−(1/2)|< /24,soδ= min(1/4, /24).

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