2023重慶專升本數(shù)學(xué)試題庫(kù)

本文格式為Word版，下載可任意編輯——2023重慶專升本數(shù)學(xué)試題庫(kù)習(xí)題1?5

1?計(jì)算以下極限?

2x(1)lim?5?x?2x?322x?52解lim??5??9?x?2x?32?32x(2)lim2?3?x?3x?12(3)2?3x?3解lim2??0?2x?3x?1(3)?122x?1?(3)limx?x?1x2?12(x?1)2x?2x?1?lim?limx?1?0?0?解lim2x?1x?1(x?1)(x?1)x?1x?12x?1324x?2x?x?(4)limx?03x2?2x3224x?2x?x4x?2x?1?1??lim解limx?03x2?2xx?03x?22(x?h)2?x2(5)lim?

h?0h222(x?h)2?x2x?2hx?h?x解lim?lim?lim(2x?h)?2x?h?0h?0h?0hh(6)lim(2?1?1)?

x??xx21?lim1?2?解lim(2?1?1)?2?limx??x??xx??x2xx22x(7)lim2?1?x??2x?x?11?122?1?limx?1?解limx2x??2x?x?1x??2?1?122xx2xx?(8)lim4?2x??x?3x?12x?0(分子次數(shù)低于分母次數(shù)?極限為零)?解lim4x?2x??x?3x?11?1223x?xxx?lim?0?或lim4x??x?3x2?1x??211?2?4xx2(9)limx2?6x?8?

x?4x?5x?42(x?2)(x?4)x?limx?2?4?2?2?解lim2?6x?8?limx?4x?5x?4x?4(x?1)(x?4)x?4x?14?13(10)lim(1?1)(2?1)?2x??xx1)?lim(2?1)?1?2?2?解lim(1?1)(2?1)?lim(1?x??xx2x??xx??x2(11)lim(1?1?1?????1)?

n??242n1?(1)n?12解lim(1?1?1?????1)?lim?2?nn??n??12421?2(12)lim1?2?3?????(n?1)?2n??n(n?1)n1?2?3?????(n?1)2?1limn?1?1??lim解limn??n??2n??n2n2n2(n?1)(n?2)(n?3)?

n??5n3(n?1)(n?2)(n?3)1?(分子與分母的次數(shù)一致?極限為解lim3n??55n最高次項(xiàng)系數(shù)之比)?

(n?1)(n?2)(n?3)11)(1?2)(1?3)?1?或lim?lim(1?n??5n??nnn55n3(14)lim(1?33)?

x?11?x1?x(13)lim2131?x?x?3??lim(1?x)(x?2)lim(?)?lim解

x?11?x1?x3x?1(1?x)(1?x?x2)x?1(1?x)(1?x?x2)x?2??1?x?11?x?x22?計(jì)算以下極限???lim32(1)limx?2x2?

x?2(x?2)32(x?2)20x?2x???解由于lim3??0?所以limx?2(x?2)2x?2x?2x2162x(2)lim?x??2x?12x解lim??(由于分子次數(shù)高于分母次數(shù))?x??2x?1(3)lim(2x3?x?1)?

x??解lim(2x3?x?1)??(由于分子次數(shù)高于分母次數(shù))?

x??3?計(jì)算以下極限?(1)limx2sin1?

x?0x解limx2sin1?0(當(dāng)x?0時(shí)?x2是無(wú)窮小?而sin1是有界變量)?

x?0xx(2)limarctanx?

x??x解limarctanx?lim1?arctanx?0(當(dāng)x??時(shí)?1是無(wú)窮小?

x??x??xxx而arctanx是有界變量)?

習(xí)題1?6

1?計(jì)算以下極限?(1)limsin?x?

x?0x解limsin?x??limsin?x???

x?0x?0?xx(2)limtan3x?

x?0x解limtan3x?3limsin3x?1?3?

x?0x?03xxcos3x(3)limsin2x?

x?0sin5x解limsin2x?limsin2x?5x?2?2?

x?0sin5xx?02xsin5x55(4)limxcotx?

x?0解limxcotx?limx?cosx?limx?limcosx?1?

x?0x?0sinxx?0sinxx?0(5)lim1?cos2x?

x?0xsinx2x?lim2sin2x?2lim(sinx)2?2?解lim1?cos2x?lim1?cosx?0xsinxx?0x?0x?0xx2x221?cos2x2sinx?2limsinx?2?或lim?limx?0xsinxx?0xsinxx?0x(6)lim2nsinx(x為不等于零的常數(shù))?nn??2sinxnx2n?x?x?解lim2sinn?limn??2n??x2n2?計(jì)算以下極限?

1(1)lim(1?x)x?x?01lim(1?x)xx?01(?1)?lim[1?(?x)](?x)x?01?{lim[1?(?x)](?x)}?1?e?1?x?0解

1(2)lim(1?2x)x?x?011?21x2x2?[lim(1?2x)x]2?e2?解lim(1?2x)?lim(1?2x)x?0x?0x?0

(3)lim(1?x)2x?x??x解lim(1?x)2x?[lim(1?1)x]2?e2?

x??xx??x(4)lim(1?1)kx(k為正整數(shù))?

x??x解lim(1?1)kx?lim(1?1)(?x)(?k)?e?k?

x??x??x?x4?利用等價(jià)無(wú)窮小的性質(zhì)?求以下極限?(1)limtan3x?

x?02xsin(xn)(2)lim(n?m為正整數(shù))?

x?0(sinx)msinx?(3)limtanx?x?0sin3x(4)limsinx?tanx?x?0(31?x2?1)(1?sinx?1)解(1)limtan3x?lim3x?3?

x?02xx?02x21n?mn??sin(xn)x?lim??0n?m?(2)limx?0(sinx)mx?0xm???n?m1x2sinx(1?1)sinx?lim1?cosx?lim21?cosx?lim?(3)limtanx?x?0x?0x?0cosxsin2xx?0x2cosx2sin3xsin3x(4)由于

2xsinx?tanx?tanx(coxs?1)??2tanxsin~?2x?(x)2??1x3(x?0)?22231?x2?1?x21x2(x?0)?~3(1?x2)2?31?x2?13sinx~sinx~x(x?0)?1?sinx?11?sinx?1??1x3sinx?tanx所以lim3?lim2??3?

x?0(1?x2?1)(1?sinx?1)x?012x?x33?求以下極限?(1)limx2?2x?5?

x?0(2)lim(sin2x)3?

x??4(3)limln(2cos2x)?

x??6(4)limx?1?1?

x?0x(5)lim5x?4?x?

x?1x?1(6)limsinx?sina?

x?ax?a

(7)lim(x2?x?x2?x)?

x???解(1)由于函數(shù)f(x)?x2?2x?5是初等函數(shù)?f(x)在點(diǎn)x?0有定義?所以limx2?2x?5?f(0)?02?2?0?5?5?

x?0(2)由于函數(shù)f(x)?(sin2x)3是初等函數(shù)?f(x)在點(diǎn)x??有定義?所以

4nx)3?f(?)?(si2n??)3?1?lim(si244x??4(3)由于函數(shù)f(x)?ln(2cos2x)是初等函數(shù)?f(x)在點(diǎn)x??有定義?所以

6limln2(co2sx)?f(?)?ln2(co2s??)?0?

66x??6(x?1?1)(x?1?1)x(4)limx?1?1?lim?limx?0x?0x?0xx(x?1?1)x(x?1?1)?limx?011??1?

x?1?10?1?12(5x?4?x)(5x?4?x)(5)lim5x?4?x?lim

x?1x?1x?1(x?1)(5x?4?x)?lim

2cosx?asinx?a22(6)limsinx?sina?limx?ax?ax?ax?a444x?4?lim??2?x?1(x?1)(5x?4?x)x?15x?4?x5?1?4?1x?asin2?coa?limcosx?a?lims?a?1?coas?

x?a2x?ax?a22(x2?x?x2?x)(x2?x?x2?x)(7)lim(x?x?x?x)?lim

22x???x???(x?x?x?x)22?lim2x2?lim?1?

x???(x2?x?x2?x)x???11(1??1?)xx

4?求以下極限?(1)limx??1ex?

(2)limlnsinx?

x?0x(3)lim(1?1)2?x??x(4)lim(1?3tan2x)cotx?

x?02xx?13?x(5)lim()2?x??6?x(6)lim1?tanx?1?sinx?

x?0x1?sin2x?x解(1)lime?ex??1xx??xlim1?e0?1?

(2)limlnsinx?ln(limsinx)?ln1?0?

x?0x?0xx(3)lim(1?1)2?lim(1?1)xx??x??xx(4)lim(1?3tanx)x?02cot2xx??12?e?e?

1x)3tan2x3?e3?

12?lim(1?3tanx?0?2?

6?x??3x?1x?13?x?32(5)()?(1?)?36?x2?由于6?x6?x6?x?3lim(1?)?3?e?lim?3?x?1??3?

x??x??6?x26?x2x?1?33?x2所以lim()?e2?

x??6?x(1?tanx?1?sinx)(1?sin2x?1)1?tanx?1?sinx?lim(6)lim

22x?0x?0x1?sinx?xx(1?sinx?1)(1?tanx?1?sinx)2xtanx?2sin(tanx?sinx)(1?sinx?1)2?lim?lim22x?0xsinx(1?tanx?1?sinx)x?0xsinx22x?(x)22?1??limx?02x38?求以下極限?

21?(1)limx?x?x?1(x?1)2(2)limx(x2?1?x)?

x???(3)lim(2x?3)x?1?

x??2x?1sinx?(4)limtanx?3x?0xxxx1a?b?c(5)lim()x(a?0?b?0?c?0)?x?03(6)lim(sinx)tanx?

x??22(x?1)2x1???解(1)由于lim2?0?所以lim?x?x?1(x?1)2x?1x?x?1x(x2?1?x)(x2?1?x)(2)limx(x?1?x)?lim

2x???x???(x?1?x)2?limx???x1?lim?1?

x2?1?xx???1?1?12x22x?1?12x?322x?1x?1(3)lim()?lim(1?)?lim(1?)22

x??2x?1x??x??2x?12x?12x?11222?lim(1?)(1?)2

x??2x?12x?12x?11222?lim(1?)?lim(1?)2?e?

x??x??2x?12x?1sinx(1?1)sinx(1?cosx)sinx?limcosx?lim(4)limtanx?

x?0x?0x?0x3x3x3cosxsinx?2sin2x2x?(x)22?lim2?1?limx?0x?02x3cosxx3(提醒?用等價(jià)無(wú)窮小換)?

(5)lim(a?bx?03xx?cx)1x?lim(1?a?b?cx?033xxx3ax?bx?cx?3??3)ax?bx?cx?33x?

由于

xxxlim(1?a?b?c?3)ax?bx?cx?3?e?

x?03xxxxxxa?b?c?31a?1b?1clim?lim(???1)x?03x3x?0xxx?1[lnalim1?lnblim1?lnclim1]

t?0ln(1?t)u?0ln(1?u)v?0ln13(?v)?1(lna?lnb?lnc)?ln3abc?3xxx13a?b?c所以lim()x?elnabc?3abc?x?03提醒?求極限過(guò)程中作了變換ax?1?t?bx?1?u?cx?1?v?

(6)lim(sinx)x??2tanx?x??1?(sinx?1)tanxsin?lim[1?(sinx?1)]x?12由于

x??1lim[1?(sinx?1)]sixn?12?e?

lim(sinx?1)tanx?limx??2sinx(sixn?1)

?coxsx?2sinx(sin2x?1)xcoxs?0??lim??limsinx?1x??cosx(sinx?1)x??sin22所以lim(sixn)tanx?e0?1?

x??21?研究以下函數(shù)的連續(xù)性?并畫出函數(shù)的圖形?

?x20?x?1f(x)?(1)?2?x1?x?2?

?解已知多項(xiàng)式函數(shù)是連續(xù)函數(shù)?所以函數(shù)f(x)在[0?1)和(1?2]內(nèi)是連續(xù)的?在x?1處?由于f(1)?1?并且

lim?f(x)?lim?x2?1?lim?f(x)?lim?(2?x)?1?

x?1x?1x?1x?1所以limf(x)?1?從而函數(shù)f(x)在x?1處是連續(xù)的?

x?1綜上所述，函數(shù)f(x)在[0?2]上是連續(xù)函數(shù)?

?x?1?x?1(2)f(x)???

1|x|?1?解只需考察函數(shù)在x??1和x?1處的連續(xù)性?

在x??1處?由于