微积分(III)期中考试参考答案.pdf

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1、‡‡‡ÈÈÈ©©©(III)ÏÏÏ¥¥¥•••ÁÁÁëëë•••‰‰‰YYY˜!ÀJK"1.D.2.D.3.B.4.D.5.D!W˜K"p1nn5x22x33+4n1.lim=:2.limsin2=0.x!13x+12n!142n3n+1ln(1+f(x)sin2x)3.®•limf(x)•3§…lim=2,Klimf(x)=3.x!0x!0arcsin3xx!0f(lnx)14.®•f(0)=1;f0(0)=1;,Klim=1:x!11x5.XJd[f(sinx)]=sin2x;Kf0(1)=1dx2n!OŽK"1.ppppp1x3(1x3)(1x+3)((3x)2

2、23x+4)limp=limppppx!82+3xx!8(2+3x)(3x)223x+4)(1x+3)pp(x+8)(3x)223x+4)=limpx!8(x+8)(1x+3)pp(3x)223x+4=limpx!81x+3(2)22(2)+4=p9+3=22.x2xx21xlimlimx=lim1+1x21=ex!1x21=e0=1x!1x21x!1x21xxx3.limcoscos


cosn!1242nx=0ž§xxxlimcoscos


cos=lim1=1:n!1242nn!1x6=0ž§cosxcosx


cosx

3、sinxxxx242n2nlimcoscos


cos=limn!1242nn!1sinx2nsinxsinxsinx=lim=lim=:n!12nsinxn!12n
xx2n2nnþµ8

4、+1x2+x2p11ln3
xdž,limx(3x3x+1)=lim=ln3.x!+1x!+1x2+x2.8>>ex1;x<0>:bsin(x1)+1;x1¦a;bŠ§¦f(x)3x=0Úx=1?Œ")µdf(x)3x=0?ŒŒ•Ù3x=0?ëY§llimf(x)=limf(x)=f(0);x!0x!0+=limex1=a;a=0:x!0´•:f(1)=1;u´0f(1+
x)f(1)1+
x1f(1)=lim=lim=1;
x!0
x
x!0
x0f(1+
x)f(1)bsin(1+
x1)+11bsin
xf+(1)=l

5、im=lim=lim=b;
x!0+
x
x!0+
x
x!0+
xÏf(x)3x=1?Œ§kf0(1)=f0(1),=b=1:yŒ•µa=0;b=1ž§f(x)3x=0Úx=1?+Œ§Ïa=0;b=1ž§f(x)3x=0Úx=1?Œ"x2n+113.?Øf(x)=limkÃmä:§ek§•ÑÙa."n!1x2n+1112nxx2n1)µjxj<1ž§limx=0;džf(x)=1;jxj>1ž,f(x)=lim=;jxj=1ž,n!1n!11+1xx2nf(x)=1;jxj=1ž,f(x)=0.u´8>>1;x<1>>x><1;1x<1f(x)=>>>>

6、0;x=1>:1;x>1x´•µx=1•Ù•˜mä:§…•amä:"Ê!A^K"(1)A(1+r)8;(2)A(1+r)128;(3)limA(1+r)8n=Ae8r:12n!1n8!y²K"®•A>0;x>0;x=1x+A;

7、^üNk.7k4•ny²µfxgÂñ¶¿¦Ñlimx.1n+12nxnnn!1nqpy²µx=1x+A;dux>0;x1
2x
A=A>0:22nxn1221x12pe^êÆ8B{y²µn2ž§ðkxnA:n=2ž§cã®y"pqpbn=kž§xA>0;Kx=1x+A1
2x
A=A;kk+12kxk2kxkppén2§ð

8、kxA:n2ž§xn+1=11+A;duxA;Ïdx2A.lnxn2x2nnnkxn+111+A1;=l1‘m©§xüN4~"xn2AnÏ•Kk•‘ƒ§ØUCêÂñ5§¤±Ø”bxnl1‘m©"pdc㌕¶fxg1üN4~§…Axx:¤±düNk.7k4•nŒ•µfxgÂñ"nn=2n2n1Applimx=a;Klimx=limx+;¤±a=1a+