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高数第一册第一章习题1.13.(1)(,1)(1,)(2){|1,}1(1,1)(1,)(3)(1,1)xxxR或(-,)(4)22903[3,1)(1,3]10xxxxx≤≤3>>1或<-12222(5)(,3)(6)sin0,,()241(7)114(1),11(1)3xxkxkkzxxxxxx(8)0ln0xxxxx>>0>1>>1(9)[1,2](10)21()xxxfxxxxxxxx-1<00≤≤10即0<<1<0和0<≤2e1≤≤27.(1)(2)(3)(4)(5)奇函数偶函数偶函数偶函数非奇非偶(6)2233()(1)(1)()fxxxfx偶函数(7)11()lnln()11xxfxfxxx奇函数)(8)2112()()2112xxxxfxfx奇函数(9)()sincosfxxx非奇非偶13.(1)22(())(2)24,(())2,xxxxfxffxxR(2)11(())(0,1)111xffxxxx(3)32221,()(1)3(1)256()56(1)(1)5(1)6xtftttttfxxxfxxx则x=t-1,或:14.2222(1)(0)0.(2)10,1111111(3)01(4)1lg,lg1,lg1,.1(5)11()(6)1log(16)yxxyxxyxyyxyxxyyyxxyxyyxxyxxxyxxxx反函数定义域定义域反函数x=,x-1=,x=1+反函数y,定义域反函数定义域x>0反函数,定义域(x)-<<1反函数(16)<<+习题1.22。(1)0,解不等式sinsin10nnnnn,得1n(2)0,解不等式111012nnnnn,得214n(3)0,解不等式10110.99...9111010nnn,得1lgn当0.0001时,41lg410n(4)0,解不等式222233232412122(21)2(28)2(2)nnnnnnnn,得122nN3.证:*nlim0,,nxaNNnN,有nxa。于是nnxaxa*0,,NNnN,有nxa,即nlimnxa4.(1)2nn211000lim1000lim0111nnnn(2)1111nnn(1)/(1)111limlimlim(1)/(1)111nnnnaababbbaba(3)nn1limlim(1)11n(4)222233nnn13(21)(41)4limlimlim33nnnnn……(公式法,利用平方和公式)(5)nn2()113limlim23(2)()33nn(6)23135212222nnnx……23135721212222nnnx……1231135232222nnnx……123122222221111121(1)1(1)12322222222nnnnnnxx…………1nlim(2)3nnxx(7)222222nnn132435(3)(1)(2)(1)(1)11limlimlim234(2)(1)22nnnnnnnnnnn……6.(1)22222222+11111111n+10=0+12nnnnnnnnn个()()故222111lim0+12nnnn()()(2)33333339327100!1234(1)22nnnnnn(3)设1,0abb,则222200(1)(1)(1)(1)1...22nnnnnnnnnnabnbnbbb(4)21111111111(1)1nnnnnnn则211110(1)(1)(1)01nnnnnnn7(1)2111313131nnx……111031nnnxx,nx单增。211113111()3332332nnnx……,有上界。(2)22211111211nxn……1210(1)1nnxxn,nx单增。22211111111(1)21212(1)nxnnnn…………,有上界。(3)11111111(1)(1)(1)242111(1)(1)(1)222111201111111(1)(1)(1)1124224221111(4)(1)(1)(1)242111(1)(1)(1)242nnnnnnnnnnnnnnnnnxxxxxxxexx…………因为又∵∴为增函数又lnlnln?…+ln?…有上界∴有上界…………11(1)2nnx减函数10.444416162222242222200022000221limlim44(2)(2)2sinsin2sinsincoscos1122224limlimlim2cos4222coscossin2sin:limlimlimcos2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxorxx(1)(2)222110000cos2sin4cos11(100)222ln(1)(3)limlimln(1)lnlim(1)1:ln(1)(4)limln(1)limln(1)11ln(1)(5)lim(1.log)limlnxxxxxxxxxxaxyxxxxxxxxxxxxxayayxxa或利用∽设(y+1)=用(3)的结论0011limln11ln(1)log(1)ln1(6)limlimlnyaxbxbbbxbxbayyyyaaaaaaaxbxb12.22(1),2,22221(1)34132111211121234122221111111(1)()()22223111(1)(2)nmnnnpnmNnmmmxxmmmmmmmmmmnnnxxnn故得:………………取故发散(2)法一:单调有界数列有极限判定{x}单增,且x有上界。法二:柯西准则n,pN,有21()111111(1)(1)(2)(1)()10,npnnpnnnnnpnpnnpnNNnxx…………,当时,pN,有故收敛。13.00:000:000:000:000:000:0xxaxaxxxaxxaMxxaMxxa(1),,当时,有f(x)-b(2),,当时,有f(x)-b(3),,当时,有f(x)-b(4),,当时,有f(x)(5),,当时,有f(x)(6),,当时,有f00:00:MxxXxxX(x)(7),,当时,有f(x)-b(8),,当时,有f(x)-b14.222222222(1)0,4225221,133252min{1,},0,0.:02,5,5,4,lim431311(3)0,,212122(21)2442222(7)0,2,222max4,xxxxxxxxxxxxxxxxxxxxxxxxxxxx00000000(9)0,,xxxxxxxxxxxxx17..1324122203341(1)1(2)lim1011(3)lim03111(4)0.sin10(1)(5)lim0(1)(1)(6)lim(33)3(12)(12)11(7)limlim4(3)(12)122(2)(4)(8)lim(4)xxxxxxxxxxxxxxxxxxxaaxxaxxxxxxxx因为有界,121212011231243(123)(1)()(9)lim(1)()(10)lim(6116)6(11)lim()()1(1)122(4)(2)(12)lim(2mmnnxxnnnnxxxxxxmxxxnxxxxxxnnnxxx……+1或用等价无穷小转化或罗比达法则+?…+1+?…+1)+(+?…+1+?…+1?………212020120021202012121)(24)2lim()(13)lim()0()mmxmmmmxmmnmnmnmxxaaaaaxxxmnbbbbbxxxaaaaxxxmnbbbbxxxxmn……时…………时……时18.02300220200sin(1)limsin2sinsin(1cos)sin112(2)limlimcos4cos2()22sin12(3)lim4sin2()2(4),0sinlimsinsin()sinlimlimsinsin()xxxxtxtaxbaaabxbbxxxxxxxxxxxxxxttmtmmxmmtntnnxnnt(mn奇偶相同0sinlimsintmtmntn)(mn奇偶不同)1900002220000235200sinsin(1)limlim1sin~sin(11)(2)limlim0(11)211sin1(3)limlim(1)0sin(sin)sinsin46(4)limlim(4xxxxxxxxxxxxxxxxxxxxxoxxxxxtgxxtgxxoxxxxxxx等价无穷小高阶无穷小高阶无穷小323526)4046()xxxxxOx同阶无穷小202lim1lim[(1)(1)]01,1.1xxxxaxbaxbabx原式2122lim()lim(35)lim(35)1xxxafxabxbxax当a=-5,b=0时,f(x)为无穷小量,当a为任意数,b0时,f(x)为无穷大量习题1.35.11000000002200000000sinsin()lim()lim1,lim()lim||1sinsinlim()lim1,lim()lim1xxxxxxxxxxfxfx
本文标题:川大版高等数学(第一册)部分课后题详细答案
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