大一上-高數(shù)二課堂

一.2ex1設(shè)函數(shù)fx)

xx2Bx x問1).AB為何值時(shí),fx)在點(diǎn)x0處可導(dǎo) fx)是否連續(xù)解:(1).可導(dǎo)必連f(00)2A,f(00) f(0)由f(00)f(00)f(0),得A f(0)

2ex1

2(ex

f(0)

x2Bx1

x0 由f(0)f(0),得BA1,B2ex(2).f(x)

xx2x x2f(00)lim2exf(00)lim(2x2)f(0)由f(00f(00f得fx)在點(diǎn)x0處連續(xù)又f(x)在(,0)及 連續(xù) 故f(x)連續(xù) 3fx)在x0連續(xù)limfx)fx0xf(x00)f(x00)f(x0fx)在x可導(dǎo)

fxfx00 x x0f(x0)f(x04例2對一切實(shí) x,

f(x

sin2x(1)fx)在x；2fx)在x處可導(dǎo)將x代入fx)sin2xf(sin2xf(x)sin2limsin2xlimf(x)f(

f(x)x即fx)在x連續(xù)由于f(

sin2xf(x)f()sin2xx x x

sin2xtx

sin2 xx

t 準(zhǔn)則limf(x)f()0即f( ) x 二、函數(shù)的連續(xù)性與間斷點(diǎn)已知函例 已知f(x)

x4ax

x1在x1求a，b.

(x1)(x

x解：Qfx)在x1處連續(xù)

x4ax

f(1)x1(x1)(xlim(x4axb)1ab ba6

x4axa

(x1)(x3x2x1

(x1)(x

(x1)(x4a3

a2,b7判別函數(shù)的間斷 (x (x3)2例 函數(shù)f(x) 2x的間斷(x1)(x1)(xx1x1x3為跳躍間斷x0

xx0為可去間斷xxxx0為跳躍間斷x1,x0為無窮間斷x1 x0為震蕩間斷x以上x可換成xaxa等(a是常數(shù)已知間例設(shè)fx)

(exb)(xb)(xa)(x

試確

a,

的值，使

f(x

有無間斷點(diǎn)x0可去間斷點(diǎn)x解：Qx0為fx)的無窮間斷l(xiāng)imxax10并且lim(exbxb)x0 x0a0b0且b又由于x1是fx)的可去間斷所以lim(exbxb(eb)(1bbe或b綜合上述，a0,b 證明函例6設(shè)fx)在x0連續(xù)，f(00又對一切x1x2fx1x2fx1fx2試證明：fx)在(,)上在fx1x2fx1fx2取x1x2 則f(0)f(0)f(0)0f(0)對x(,)yf(xx)f(f(x)f(x)f(x)f(x)[f(x)limylimf(x)[f(x)1]f(x)[f(0)1]x0 x0fx)在(,)三、閉區(qū)間上三、閉區(qū)間上連續(xù)函數(shù)的性例7設(shè)fx)在[上連續(xù)，f(1)1試證明：使f()令gxxfx則gx)在[上連g(0)1 g(1)f(1)1由零值定理(0,1使g(即f()由于0，故f( 例 若函數(shù)f(x)在[0,1]上連續(xù)，證明方程f2(x)1x證明令gxx2[f2x1則gx)在[g(0)10,g(1)f2(1)g(0)g(1)由零值定理，g(即為方程在（內(nèi)的解例 證明：直線y2x1與曲線y2x至少有一個(gè)交點(diǎn)證明令fx2x12fx)在[上連續(xù)，且f(0)20,f(1)12由零值定理，(0,1使得f(即直線y2x1與曲線y2x用介值例 設(shè)f(x)在(a,b)內(nèi)連續(xù)，ax1x2b,t10,t2證明(ab使t1fx1t2fx2(t1t2f(證明Qfx)在x1x2(ab)由最值定理fx)在x1x2上存在最大值Mm mf(x1)M mf(x2)(t1t2)m t1f(x1)t2f(x2)(t1t2mt1f(x1)t2f(x2)Mt1t2x1x2(abf()t1f(x1)t2f(x2t1 t1f(x1)t2f(x2)(t1t2)f(例

設(shè)fx)在[0n]f(0)f(nn[0n1]使f(1f(證明令Fxfx1f則Fx)在[0n1]上連續(xù)，且F(0f(1fF(1) f(2)fF(2) f(3)f(2F(n1)f(n)f(n于是F(0F(1F(n1f(nf(0設(shè)F(x)在[0，n1]上最大值為M，最小值為 m1[F(0)F(1)...F(n1)]M由介值定理 [0,n1],使F()1[F(0)F(1)...F(n1)]n f(1)f(練習(xí)一/(1)當(dāng)x時(shí)fx

x2x

分析：Qlime x10

lime x1x1

x2xx2

eeee

lim(xx1

x10

x

選易犯錯(cuò)誤：沒有看 x

x1左右極限不同誤認(rèn)為ex ,錯(cuò)選C (2)當(dāng)x0時(shí)，下列四個(gè)無窮小中哪一個(gè)是比其它 1x2(A)x2 (B)1cos (D)tanx1x2

1cosx~1x21x1(1x2)121~1x2而tanxsinxtanx(1cosx~x1x21 x0tanxsinx是x的三階無窮故選當(dāng)xtanxsinx是關(guān)于基本無x( 一階無窮?。–）三階無窮小

（D）limx

(x

u limtanusinulimu2kcu k故

u（設(shè)f(x)(1cosx)ln(1x2 g(x)xsin(xn2h(x)e

其中n為正整數(shù)。已知x時(shí)，f(x)是比g(x)高階的無窮g(x)是比h(高階的無窮小，則必有n ( (A) (B) (C)

(D)f(x)(1cosx)ln(1x2)~1x22h(x)e 1~xg(x)~xn1據(jù)題意，2n1 即1n3 n 故選 二、填(2)設(shè)函數(shù)fx

2x3x

或lnx，則當(dāng)x 2

,0,

f

是無窮大解m2x3lnx lim2x3lnxx13x，

x33x2lim3x2

lim2x3lnxx22x ln3

x23x3lim3x2

lim2x3lnxx

2x ln

x

3xlim3x2

lim2x3lnxx2x

ln

x3x五、求下列各題中的常數(shù)A和五、求下列各題中的常數(shù)A和（2）x2x24x~A(xx2xlix2xkx2 A(xk

x244 x2x24xx2

A(x42 lim(x2)2k42x2422k42

kA1 十一、求下列極(1)limln(2x

(等價(jià)無窮小代

x

(x)0,ln(1(x))~(x)limln(12x

x23lim2(x3) x x

ecos

(等價(jià)無窮小代x2

2x

(x)0,e(x)1~(x)解：原極

cosx2

2(x2sin(lim 1x2

2( 2十四、求下列各題中的常數(shù)A，B和正數(shù)k，(1x212xB1~Ax2)k6x2x12xB

6x2

x

lim12xB6x

x

A(x

x2x2(6x)A(x

x36x212x 16A

(x由limx36x212xB)0Bx代入上式得

x36x212xk16A

(x (x 116Ax2(x2)k

求得：A

，k B6xx24~A(x3)k求常數(shù)AB和正數(shù)k.BB6xx2

B故BB

4)

4B6B6xx

76xx2

~(x3)2776xx2A1,k a(a1)若a是正數(shù),x1 a(aa(a1)(n1,2,L),證明數(shù)列xn收斂lim解:xn1xna(a(a1)xna(a1)

a(a1)a(a1)a(a1)可知xn