高数(上)习题及答案(极限).pdf
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高数(上)习题及答案(极限).pdf
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1复习题
(一)一、定义域
(1)函数的定义域()2lg31+=xxy
(2)函数的定义域1412+=xxy(3)函数的定义域为0,1,则定义域为()xf()+=4141xfxfxg二、求极限
(1)+xxxxx1sin2sin3lim00312limsinsin22xxxxx=+00312limlimsinsin22xxxxxx=+003lim2sin2lim2xxxx=32=()lim0mmnnxaxaaxa11limmnxamxnx=limmnxamxn=mnman=xxx+21lim222lim1xxx=+2222lim1xxx=+2e=xxxlnlim1lim0xx=4586lim224+xxxxx42lim1xxx=23=xxxxxtan2sinlim000sin2n1lim2lim2cosxxxsixxxx=000sin2n12limlimlim2cosxxxxsixxxx=1=211lim20+xxx()20lim11xx+=+1sinlim0xxex0slim1xxcoxe=;exxex1lnlim11limxexe=;()11lim22+xxx222lim011xxx=+;xxx1sinlim1sinlim11xxx=xxx211lim221lim1xxex=xxx3cos5coslim225sin5lim3sin3xxx=53=3022sinlimln0sinlimxxxxxxxex+=20cossinlim2sinxxxxxxxe+=0sin2lim21xxxxe+=()11111lnlimlimln1ln1xxxxxxxx=111lim1lnxxxxx=+11lim1lnxxxxx=+11lim2lnxx=+12=111lim0xxex()01lim1xxxexxe=01lim1xxxxeexe=+0lim2xxxxeexe=+01lim2xx=+12=1131232limlim12112xxxxxxxx+=+411312lim112xxxxx+=+2323122331122lim111122xxxxxxx+=+e=解法2:
原式12lim121xxx+=+211222lim121xxx+=+2112222lim112121xxxx+=+e=解法1:
xxxx30sin1sin1lim()30sinlimsin1sin1xxxxxx=+301sinlim2sinxxxx=2011coslim23sincosxxxx=220112lim23xxx=112=解法2:
xxxx30sin1sin1lim3011sin22limxxxx+=301sinlim2xxxx+=5201cos1lim23xxx+=220112lim23xxx=112=()221arctan12limlim111sincosxxxxxxx+=洛比达法则22lim1xxx+=+1=不存在xxx1sinarctan2lim()xxxtan2sinlim,tansinxyx=解:
令tanlnsinyxx=则ln22limlimtanlnsinxxyxx=ln2lnsinlim1tanxxx=222cossinlimsectanxxxxx=22coslimsinsinxxxx=21limsin22xx=0=,02limlnxye=ln2lim1xy=6解:
sinsinln00limlimxxxxxxe+=0limsinlnxxxe+=00lnlimsinlnlim1sinxxxxxx+=021limcossinxxxx+=20sinlimcosxxxx+=0limsinxx+=0=sin0lim1xxx+=xxx3tan6sinlim6,6tx=解:
令06limsintan3limsincot36txxxtt=则0cos3limsinsin3tttt=0sinlimsin3ttt=13=xxxx5sinsin3sinlim000sin3sin2sin2coslimlimsin5sin5xxxxxxxx=解:
02sin2limsin5xxx=45=7;+2sin12coslim220xxxx2222200212limcoslimcossinsin22xxxxxxxxx+=+解:
222002limcoslimsin2xxxxxx=+2220022limcoslim2sin2xxxxxx=+0=30sin1tan1limxxxx+330011tansin1tan1sin22limlimxxxxxxxx+=解:
301tansinlim2xxxx=()30sin1cos1lim2cosxxxxx=2011coslim2xxx=01sinlim22xxx=14=;()xxxx2coscos11lim208()201lim1coscos2xxxx解:
;()1ln10lim+xexx()()1lnln1ln100limlimxxxeexxxe+=解:
()0lnlimln1xxxee+=01limxxxexee+=0limxxxxeexee+=01lim1xxe+=e=
(2),则()32lim=+xxkxkx=k(a)(b)(c)(dddd)2e21e23ln31221lim321kxkkxxkkxkx+=分析:
32133ln33kkkeeke=(12)数列的极限是()nnnxncos+=(aaaa)1(b)-1(c)0(d)不存在cos1limlimlim1cos1nnnnnnxnnn+=+=分析:
(16)()()()()=+3321limnnnnn(a)0(b)1(c)3(d)69()()()3123123limlim1111nnnnnnnnn+=+=分析:
(18)()=xxx3sin5sinlim(a)(b)-1(c)1(dddd)3435sin55cos55limlimsin33cos33xxxxxx=分析:
(22)()=+xxxxxsin31lim2(a)-1(bbbb)-2(c)1(d)22211313limlim2sinsin1xxxxxxxxx+=+分析:
(23)若,则()()51sinlim21=+xbaxxx(a)(b)3,5=ba6,7=ba(cccc)(d)4,3=ba1,0=ba()()2211lim5lim01sin1xxxaxbxaxbbax+=+=分析:
由,可知,即()()()1111lim5sin1xxxabax+=将代入()1lim153xxaa+=得,即(26)()=+112cos1lim22xxxxxx(a)0(b)1(cccc)2(d)3102222121121limcoslimcoslim11xxxxxxxxxxxxx+=+分析:
221121limcoslim11xxxxxxx+=+2=(28)如果,则()322sin3lim0=xmxx=m(a)1(b)2(c)4/9(d)9/4003sin23sin2324limlim2323239xxmmxmmxmmmxmx=分析:
(29)()=+22221limnnnnn(a)0(b)(cccc)1/2(d)1()2222111212limlim2nnnnnnnnn+=分析:
(30)若,则()2134lim2=+baxxxx(a)(bbbb)6,2=ba2,4=ba(c)(d)1,3=ba2,0=ba243lim24041xxaxbaax+=+=分析:
由,可知()43lim24221xbxbbbx+=+=(31)的值为()xxxxsin1sinlim20(a)1(b)(c)不存在(dddd)01120011sinsin0limlim0sinsin1xxxxxxxxx=分析:
(32)()=222sinlimxmxx(aaaa)0(b)(c)(d)2m22m2222sin1limlimsin022xxmxmxxx=分析:
三、函数连续(5)已知函数在连续,则()()()=0,0,211xbxxxfx0=x=b()()0lim0xfxf=分析:
函数在一点连续的充分必要条件是而()()1200limlim12xxxfxxe=()0fb=2be=(6)已知函数在连续,则()()()=+=0,0,21lnxbxxxxg0=x=b解:
()()()()110000ln12limlimlimln12lnlim122xxxxxxxfxxxx+=+=+=()0fb=2b=(11)已知函数在内连续,则()()=00,2sinxaxxxxf()+,=a12解:
,()()sin2,00xxfxxax=+=要使在,内连续sin2xxx当0时,是连续函数()sin2,00xxfxxxax=只要使在=0处连续()()0lim0xfxf=即只要使0sin2lim222xxaax=(20)已知函数在连续,则()()=+=11,113xaxxxxxf1=x=a(aaaa)2(b)-2(c)1(d)-1()31,111xxfxxxax=+=分析:
要使有意义()()1lim1xfxf=须使311lim13121xxaaax=+=+=即(34)若在连续,则()()+=01sin00sin1xbxxxaxxxxf0=x(a)(bbbb)1,1=ba1,1=ba(c)(d)1,1=ba1,1=ba13()1sin0001sin0xxxfxaxxxbxx分析:
要使在处连续()()()00limlim0xxfxfxf+=只要使0011limsinlimsinxxxxbaxx+=+=只要使即a=1,b=1(36)已知在连续,则()()+=02sin02xxbxxbxaxf0=x(a)(b)1,0=ba0,1=ba(c)(d)ba2=ab2=()200sinlimlim2xxbxabxax+=由题意,只要使()200limsinlim22xxabxabbxbaabx+=四、无穷小
(1)函数在哪个变化过程中是无穷小量()11=xey(a)(b)(cccc)(d)0+x0xx1+x
(2)当时,是的()0x1xex(a)高阶(b)低阶无穷小(c)同阶无穷小(dddd)等价无穷小001limlim11xxxxeex=分析:
(3)当时,是的()0xxxsin2x(a)高阶无穷小(b)低阶(cccc)同阶无穷小(d)等价无穷小14(4)已知当时,为无穷大量,则当时,下列变量必为无穷0x()xf0x小量是()(a)(bbbb)(c)(d)()xxf()xfx()xxf11+()xxf1(6)时的无穷小量()0x(aaaa)(b)(c)(d)xx1sin2xx1sin1xxsinxx2arcsin(7)曲线的垂直渐进线是()3242=xxy(a)仅有一条(b)仅有一条3=x1=x(cccc)有两条,(d)不存在3=x1=x(19)当时,与是等价无穷小,则()0x()11+kxx2sin=k(a)1(b)2(c)3(dddd)4()001112limlimsin24sin24xxkxkxkxx+=分析:
(21)当时,为()0xxy1sin=(aaaa)有界变量(b)无穷大量(c)无穷小量(d)无界变量(25)函数是()()()1ln2+=xxxf(aaaa)奇函数(b)偶函数(c)非奇非偶(d)有界函数(33)当时与等价的无穷小量是()1x()1sin2x(a)(bbbb)(c)(d)2x12x1xx15(35)当时,与比较是()1x()1sin2x1x(a)较高阶的无穷小(b)较低阶的无穷小(cccc)同阶无穷小(d)等价无穷小五、渐近线(5)曲线的垂直渐进线是()9862+=xxy(a)仅有一条(b)仅有一条9=x1=x(cccc)有两条(d)不存在9=x1=x(27)曲线的水平渐进线是()xeyx=33(a)(b)(c)(dddd)不存在0=y1=y3=y六、间段点:
六、间段点:
六、间段点:
六、间段点:
(37).111的间断点,并判断类型指出xxey=(24)函数,是函数的()()=0arctan0001xxxxexfx0=x(a)连续点(b)第一类可去间断点(c)第二类无穷间断点(dddd)第一类跳跃间断点
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