专接本数学函数与极限习题课

y=y=a

(a>0,a„

y=logaxy=sinx;y=tanx;

(a>0,a„y=cosx;y=cot y=arcsin y=arccosy=arctanx;y=arccot

sinhcosh

-= -2 -= +2

tanhx

sinh

=ex-e-cosh ex

+e-limlimsinx=xfix

limsina=1

limalim(1lim(1+1)n=nfinxfixlim(1+a)a=1lim(1+x)1lim(1+x) xfisinx~x tanx~x 1-cosx~1x22arcsinx~x arctanx~x ln(1+x)~xex-1~x ax-1~xlna (1+x)m-1~

直接法:先作辅助函数F(x)求函数ylogx-1)(16

16-x2>x-1„

x<x>x„1x2及2x y

x2,lnx

-1£x<x£12ex-1

<x£ e当-1£x0yx2˛01y则xy

y˛(0,当0x£1ylnx˛-¥,0则x=ey y˛(-¥,当1x£2y2ex-1˛2,2e],

1 2 x=1+lny y˛(2,2eex x˛(-¥,反函数y

x˛(0,

x1+lnx2

,x˛(2,2e

(-¥,1]∪(2,2e例3已知fx)x3,

x‡ f(5)

f[f(x+ x<解:f(5)f[f(10f(10=f(7)=f[f(12)12 f(sinx f(x).

sin

)=csc2x-cos2x

f(sinx

)sin

sin2

+sin2x-=(sinx

-sin\f(x)=x2- 例5fxfx12x,其中x0,x1求f t=x-x

即x

1-

f )+f(t)=

即f

)+f(x)=1- =u-

- x

1- 1-11- 1-f f(u12(u 即f

)+f(x-1)=2(x-1-

f(x)= -+ 1-+== 0中a为任意实数。nfi¥证明1a,存在正整数M使|a|£M |

Mn

nM Mn= M" M" £M 12" M+1" M Mn

e(M-

取Nmax{[eM!],M Mn<e0$NnN时，|n|£即 =nfi¥证明2a,存在正整数M使|a|£M 则|

设

M=

，

=n+1xnM当n1M时，xn+1xnxnMxn0所以xn有极限，AA=lim =

= A=nfi =0 =0

nfi¥n+

lim|

|=0nfi

=0=

nfi¥nfi¥limf(x)=Alim|f(x)|=|Axfi xfilimf(x)= lim|f(x)|=0.xfi xfikn例7：证明：lim =0（a>1,k˛Nknnfi¥a0a1an=(1+l)n=1+la+"+n(n-1)"(n-k)lk+1+"+(k+>n(n-1)"(n-k)(k+

(nk时因 0<

(k+

n(n-1)"(n-k+1)(n-k=(k+

n(1-1)"(1-k-1)(1-k

nfi

fin\ =nnfi¥8对于数列xnx2kfia(kfi¥x2k-1fia(kfi¥)，则xnfia(nfi¥e0$N1nN1时|x2k-1a|e$N2nN2时|x2ka|e取Nmax{2N112N2当n>N时 |xn-a|<所 limxn=nfik

fia(kfi¥)xnfia(nfi¥)kkk

n

则

\"k˛N*,

£a

从而对xn都有xn£a(∵k£nk,xk£k

£a,k=1,2,""e>0 $L,当k‡L时,a- =|xn-a|<e 取N=nL,则当n>N时 ∵ =xN£xn£L|xn-a|=a-xn£a-xnL=|a-xnL| limxn=nfi例10limnn1nfi1 定理证明：limx =1;xfilimxx1并求极限limxlnxfi xfi证明：(1不妨设x1 [x][x]+1<x <([x]+1)[x] [xxfi而[x][x]+1=([x][x])[x xfi[x

1\limx =1;xfi [x([x]+1)[x]=([x]+1)[x]+1

xfi

fit1(2)令x=1 则当xfi+0时,tfit11\limxx=lim(

=lim =xfi tfi

tfit\limxlnx=limlnxx=xfi xfi例11x+x+xfi

- x=limxfi=limxfi

x+xx+xx+x+ x++x

x2xx+x x+x2 无穷 有xfi

1-x令tx-t(t+tfitfi

sinp(t+1)t(t+2)sinpt t(t+ tfi

lim(1+x)a-(1+xfi

(1+x)a-

(1+x)b-= xfi0 =xfi

(1+x)a-x

-xfi

(1+x)b-x=limax-lim

=a-xfi xfi

lim4arctanx-=xfi

x-1arctanx-p4x-xfi时，arctanx4

~tan(arctanxp

tan(arctanx)-

x-tan(arctanx

) 1+tan(arctanx)tan

p=1+4x-4lim1

= =xfi

x-

xfi11+lim(1+xtx=lim(1+2x)cotxfi 1- xfi

lim(cotx2x=exfi

lim(cos

2x=exfi

sin

=

limu(x)0limv(x)¥ xfi xfi+v(x)xfix0

limv(x)u(x)=exfilimv(x)ln1+u(x)]=exfixfi

1+tan1+sin

)x3

xfi

1+tan1+sin

-1xfi

tanx-sinx]x31+sinx

=limtanx(1-cosx) xfi

1+sin

x xfi

1+sin =limtanx1-cos xfi

1

1+sin e2

n5+n7

n5+n7-

+ n5+n7-

1(

1n11n1n

7-1)lim=enfi

nfi+¥ = 11lim2nfi+¥=e

(n5-1)+lim(1nnfi1n

7-1)1nxfi1nxfi0ax-1~xln

limxsinx

=limesinxlnxlimxln=exfi =

limsinxln=exfi

xx-

=

exlnx-

=

xlnx=xfi xln

xfi

xln

xfi1xln+lim 2)n +nfi

解1:limn22lim(22)nfi nfi 所 lim(1+2nfi

)n=e2解2:lim12nfi n1

2)n

<

=1+++n<1+n++

n(n- n- 2

2

1+n

<1

n+n2

<1

n-1 2

lim1+ =e2,lim1

=nfi¥ n

nfi

n-所 lim(1+2nfi

2)n=e2

lim(1+2+"+ nfi

(n+=lim(2-1+3-1+"+n+1-nfi

(n+=lim(1-1+1-1+"+1- nfi

(n+=lim(1nfi

(n+

)=

lim

2+e4

sinxx x

xfi0 1

1+e

-

-2+e sinx

2

+

sinx

+ =

=xfi0+

41+

xfi0+

e-+ 1

4 4 12+e

sinx

2+e

sinxlim + =lim - =xfi0-1+e xfi0-1+e

qcosq"cosqnfi

解将分子、分母同乘以2sincosqcosq"cos qcosqsinq原式=

nfi 2sin=

=nfi¥2nsin 当x时n求lim(1x)(1x21x4(1x2).nfi将分子、分母同乘以因子(1-x),nnfi

(1-x)(1+x)(1+x2)(1+x4)"(1+x21-

(1-x2)(1+x2)(1+x4)"(1+x

=

(∵当

当xfi0时,下列函数是x的几阶无穷小

3x2+

3x2+k设其为xk阶无穷小,则kxfi

=C„3x2+

x2+

1-

x (1+x2xfi 故k6

xfi

x

xfi

-31+31+ 31+ 3x1 -3x3x=(31+ x)2+31+ x+3x=(31+ x)2+31+ x+故是x1阶无穷小31+tan1+tan1-sin1+tan1+tan1-sin1+tan1-sin1+tan1-sin+故是x3阶无穷小 4+x4+x24+4+x2+

-4+4+x22 2故是x2阶无穷小

ex

-cos2解1ex

-cosx=(ex

2xfi0时,ex

-1~x21-cosx

1x22可看出原式是x2阶无穷小

ex

-cos解2设其为xk阶无穷小, limexxfi0

-cosxxk

=xfi

2xex +sinxkxk-1=xfi

2ex +4x2ex +cosk(k-1)xk-故是x2阶无穷小例13设fx)x-x求limfx),limfxxfi xfinlimfx)limx-x])nnxfi xfilimf(x)=lim(x-[x])=n-(n-1)=xfi xfi

设px)是多项式,且xfi

p(x)-x3x2

=limpx)1,求pxfi解xfi

p(x)-x

可设px)x32x2axb(其中a,b为待定系数又∵limpx)xfi \p(x)=x3+2x2+ax+b~ (xfi从而 b=0,a= 故p(x)=x3+2x2+

xn(1an(1a)n求证：

1+|a=

a„nfi

a=证:当a0时

xn+1”1\limxn+1= nfi¥当a0时

(1+a)n+1+(1-lim n+1=nfi¥

nfi

(1+a)n+(1-=lim(1+nfi

1+(1-a1+1+(1-a1+

=1+当a0时

nfi

xn+1= nfi

(1+a)n+1+(1-(1+a)n+(1-(1+a)n+1+=lim(1-a)1- =1-nfi¥ (1+a)n+11-a所

1+|a a„=nfi

a=+例16. ai‡0(i=1,2,"),证明下述数列有极限+xn

(1+a1)(1+a2

+"

(1+a1)(1+a2)"(1+an(n=1,2,"证:显 xn£xn+1 即{xn}单调增,nxn=k

(1+ak)-(1+a1)"(1+ak

=1

1+n 1+1k=2

(1+

)"(1+

-(1+

)"(1+

)=1

\lim (1+a1)"(1+an

nfi¥ 2(1+x例17：设x0>0,xn+1= ,n=0,1,2,",2+nfi

x0>0,

=1

2+

>1,n=0,1,2,"又xn+12

2+

<2,n=0,1,2," 0<xn<2,有

-

=

2+

n-2+xn=(2+

)(2+xnxn+1xnxnxn-1n1,2x1x0{xn单调增加；x1£x0{xn由单调有界原理：xn的极限存在。n设limxnA,∵nfi

=2(1+xn)2+n2则A22A解得A22+2即limxn2nfi ：F0=F1=1,Fn+1=Fn+Fn-1(n‡给出的数列{FnL.Fibonacci数列证明n证明:记 =Fn+1(n=1,2," nn =n

+Fn-1=1+Fn-1=1+

F Fx >0,\x

1

<1+1= 所以数列xn既有上界 论x=1,

=F+

=2

=F2+F1=3>F2 F22x=1+1=2

=1+ xx3 =<xx3 假设： >x2(k-1) x2k+1<x2k-1 x2k+2-

=1+x

-(1+x

)=x2k-1-x2k > 2k 2k -

=1

-(1

1)=

-x2k <2k

2k

2k 2k x2kx2k-1单调减少。x2k}x2k-1}极限存在。 limx2k=A,limx2k-1=B,（式kfi kfi1x2k+1=1+

=1+

B=1+1A

A=1+B5AB1525于 limFn+1=15 例19yfxxfi¥yaxblimfx)a lim[fxaxb均存在。xfi

xfi由条件知，limfxaxbxfi\limf(x)=lim(1[f(x)-ax-b]+a+b)=xfi xfi¥ \lim[fxaxbxfilim[f(x)-ax-b]=lim[f(x)-ax]-b=xfi xfi例20a,b,1

lim(31-x3-ax-b)=xfibx解1:limxxfi

x3-1-ab

x)=3x33x3-1xfi

-a

x)=故1a0,a1,b=lim(31-x3+xfi=xfi=

3(13(1-x3)2-x31-x3+yy=-y=31-解2:由条件知，yaxb是fx)31x3a=xfi

f(x)

=limxfi

1- x3b=lim(31-x3+xfi3(13(1-x3)2-x31-x3+xfi=

1aa=b=2ab,x2-x2-x+xfi

-ax-b)=例21a,b,使xfi

x2+bx+x- =由条件知，x2bx3bx x2+bx+3b=(x-a)(x-c)=x2-(a+c)x+比较系数得，a+c=- ac=

∵

x2+bx+3b=lim(x-a)(x-c)=a-c=

xfi

x-

xfi

x-联立(1)(2)(3)式得，a4ab16b

a(1cosx

x<f(x)

1ln(b+x2)

x=x>在x=0连续,则a

,b 1-cosx~1x2 f(0-)=lima(11-cosx~1x2xfi x f(0+)=limln(b+x2)=lnxfia=1=ln2

a=2,b=

讨论fx)

x-1,

x>

的连续性解将fx

,x£2 f(x)

px,-1£x£1

x>显然fx)(-¥,-1-1,1)1,+¥内连续x1时limf(x)=lim(1-x)=2.∵ f(x)„limf(xfi- xfi-

xfi- xfi-limfxlimcospx 故fx)在x1间断xfi-

xfi- x1时limf(x)=limcospx=

∵limf(x)=limf(xfi

xfi

xfi xfilimf(x)=lim(x-1)=xfi xfi

故fx)在x1连续\fx)(-¥,-1∪(-1,+¥连续例24.

f(

x2n+1+ax2+

当a,b取何值时，=nfi

x2n+fx)在(-¥,+¥上连续。 f(

=nfi

x2n+1+ax2+bxx2n+1ax2+xx=a-b-

|x|<|x|>x=- a+b+

x= 要使fx)在(-¥,+¥fx)x1f(1)=limf(x)=limf(xfi xfif(-1)= f(x)= f(xfi- xfi-a-b=-a-b=-解得：a0,bf(x)

(1xsin x(x+1)(x-

(1+(1+x)sinx(x+1)(x-x=–1limf(x)=xfix=1limf(x)=- limf(x)=1,xfi xfix=0例26求fx)xx(1)x0nx0n1limf(x)=limx[x]= n=x0[x0]=f(x0xfi xfi即fx)在x0x0是整数时，设x0n f(x)= x[x]= n=x0[x0]=f(x0xfix0 xfix0 f(x)= x[x]=x0(n- n=0时，x0= f(x)=0=fn„0时

f(x)=x0(n-1)„f(x0x0fxf(x)

ex-

例27

(x-a)(x-x0x1,ab∵x0ex-

(x-a)(x- =

xfi

ex-

=1-b=a=0,b„x1为可去间断点,

ex-

lim(ex-b)=

xfi1x(x-b=limex=xfi xfi例28设f(x)

(-¥¥)且对任意实数x,y有fxy)fxf若f(x)在x0连续,证明f(x)x解limfxDx)limfxf(Dx)]Dxfi

Dxfi=f(x)+f=f(x+=f(

设fx)在闭区间[0,1且f(0=f(1x˛[0,1]使得f(x1)f2Fx)fx1f2Fx)在[0,1]上连续2∵F(0)

f )-f2

F(1)=f(1)2

f( 若F(0)= 则x=

f(0+1)=f2

f(1

+1)=f(1);F(2)=

则x=2 F(0)0F(1)0,2F(0)F(1)=-[f(1)-f < 由零点定理知,$x˛(0,1F(x2即f(x1)f(x成立22综上,必有一点x˛[0, 2使f(x1)f(x)成立2

设fx)在闭区间[0,1且f(0=f(14x˛[0,1]使得f(x1)f44证明Fx)fx4

1fxFx)

[0,

反证，若对任何x˛[03都有Fx)04

上 上4Fx0Fx0Fx0F(0)=f(1)-f(0)>0 F(1)

f )

f(

>0 F(2)=f(4)-f(2)>0 F(4)=f(1)-f(4)>0从 f(1)>f(0)x˛[0,1]F(x0即f(x1)f4

设fx)在闭区间[0,1且f(0=f(1证明nx˛[0,1]f(x+1)=fn例32fx在[0,1f(0f(10a(0a1),必有一点x˛[0,1],使得f(xa)f(x).证明令Fx)fxaf\F(0)=f(a)-f(0)=f(a)‡F(1-a)=f(1)-f(1-a)=-f(1-a)£F(0)f(a0,则取x0；F(1af(1a0,则取x1a˛F(0)0F(1a)0F(0)F(1a [0,1],F(x)在[0,1-a]上连续即f(xa)f例33设fx)在(a,b且xi˛(a,bi1,nx˛(a,b),f(x)

n(n+

[f(

)+2f(

)+"+nf(

证明：ax1x2"xn并设