数学分析I-14-期末复习题.pdf

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1、555êêêÆÆÆ©©©ÛÛÛI666˜˜˜!!!444•••OOOŽŽŽp1.lim(n2+nn)n!1pppnnn2.lim(1+2+


10).n!11113.lim(++


).n!11
22
3n(n+1)132n14.lim(++


).n!12222np5.lim(n1+2n+3n).n!1x216.lim().x!2x24x2(3x+1)2(x2+1)7.lim.x!1(x+1)(2x+1)3118.lim(xsin+sinx).x!0xxx2arctanx9.lim.x!12x2+1x2+ax+b10.®•li

2、m=8,¦a;b.x!2x23x+2p1+2x311.limp.x!4x2pp21+cosx12.limpx!01+x21x2sin113.limxx!0sinxp1+f(x)sinx114.®•lim=2,¦limf(x).x!0e3x1x!0x22ecosx15.lim.x!0sinxln(1+x)pp1+xsinxcosx16.lim.x!0x2e112+cosxx17.lim(()1).x!0x33118.lim(cosx)x2.x!0319.lim(1+2x)sinx.x!013x+2x20.lim().x!13

3、x+1xxxa+b+c121.lim()x;a>0;b>0;c>0.x!03222.lim(1+ln(1+x))x.x!0123.lim(cosx)ln(1+x2).x!0tanxx24.lim.x!0ln(1+x2)1125.lim().x!0xln(1+x)p126.lim(x+1+x2)lnx.x!127.lim(xx2ln(1+1)).(Vúª)x!1xx+4sinx28.lim.x!15x2cosx29.f(x)3x?Œf0(x)=1,¦00f(x0Mx)f(x0)(1)lim.x!x0Mxf(x0+2sinx)f

4、(x0)(2)lim.x!x0arctanx30.¼êf(x)÷vf(1+x)=af(x),f0(0)=b,¦f0(1).f(a+h)+f(ah)2f(a)31.f(x)3a?äkëYê,¦lim.h!0h2!!!¦¦¦ÚÚÚ‡‡‡©©©p1.y=arcsin1x2.2.y=cosnx+sinnx.p3.y=ln(x+1+x2).p4.y=arctan1+x2.5.y=tanx2+arctan2x.6.y=exsin(3x).axx27.y=a+x.dyd2y8.x=ln(1+t2);y=arctant½y=y(

5、x),¦,.dxdx2p9.y=xx+3x+x3+ln3x+ln3,¦dy.p10.y=sin1+x2,¦dy.2p11.y=ln(cos2x+1+sin4x),¦dy.112.f(x)Œ,y=xf(),¦y0;y00.x13.y=sin(2x),¦y(n).114.y=,¦y(n).x(x2)15.y=f(x2+1),¦dy.nnn!!!ØØؽ½½ÈÈÈ©©©Rx2+sin2x1.dx.x2sin2xR(x+1)22.dx.x(x2+1)Rx43.dx.1+x2R14.dx.x2+2x+2pRxtan1+x25.pdx.1+x2pR1

6、+lnx6.dx.xRp7.arctanxdx.Rxearctanx8.dx.3(1+x2)2R19.dx.3x2(1x2)2R3x210.xedx.Rlnsinx11.dx.sin2xRarctanex12.dx.e2xR13.e2x(1+tanx)2dx.Rpp14.xsinxdx.R115.dx.1+sinx+cosxRR116.xf(x)dx=arcsinx+c.¦dx.f(x)ooo!!!nnnØØØyyy²²²3n21.

7、^"N½Ây²lim=1.n!1n241112.an=1+++


+,y²liman•3.2232n

8、2n!1x2123.

9、^"½Ây²lim=.x!12x2x134.

10、^8BKy²limcosxØ•3.x!15.f3(0;+1)þ÷vf(x2)=f(x),…limf(x)=limf(x)=f(1),y²f(x)x!0+x!+1f(1);x2(0;+1).ex16.•Ñf(x)=mä:9a..x(x1)81>>>>>:1xsin+b;x>0x8.f3[0;+1)þëY,÷v0f(x)x;x2[0;+1),a10,an+1=f(an);n=1;

11、2;


.y²(1)fangÂñ.(2)liman=t,Kf(t)=t.n!19.e¼êf(x)3[a;b]þëY,…f(a)=f(b)=k,f0