利用定积分定义求极限(by汤)

利用定积分定义求极限1入门题同济7的p226Zbaf(x)dx=I=lim!0nXi=1f(i)∆xi;xi166xi做题时的公式Z10f(x)=limn!11nnXi=1f(in)Zbaf(x)dx=limn!1ban„ƒ‚…∆xnXi=1f(a+in(ba))1入门题Example1:求极限I=limn!11n+1+1n+2+


+12n.SolutionI=limn!11n11+1n+11+2n+


+11+nn!=limn!11nnXi=111+in=Z1011+xdx=ln2Example2:求极限:limn!1n!nn1n.Solutionlimn!1n!nn1n=limn!1exp1nlnn!nn=explimn!11nlnn!nn=explimn!11nln1n+ln2n+


+lnnn=explimn!11nnXi=1lnin=expZ10lnxdx=exphxlnxi10Z10dx=1eExample3:求极限:I=limn!11nnqn(n+1)(n+2)


(2n1).SolutionI=explimn!11nn1Xi=1ln1+in!=expZ10ln(1+x)dx=4eExample4:求极限:I=limn!11p12+n2+1p22+n2+1p32+n2+


+1pn2+n2by汤第1页,共9页利用定积分定义求极限1入门题.SolutionI=limn!1nXi=11pi2+n2=limn!11nnXi=11q(in)2+1=Z101px2+1dx=hln(x+px2+1)i10=ln(1+p2)Example5:求极限:limn!1112+n2+222+n2+


+nn2+n2.SolutionI=limn!1112+n2+222+n2+


+nn2+n2=limn!1nXi=1ii2+n2=limn!11nnXi=1inin
2+1=Z10x1+x2dx=12Z1011+x2d(1+x2)=12ln(1+x2)10=12ln2Example6:求极限:limn!1np1+2!+3!+


+n!n.Solution由于npn!n6np1+2!+3!+


+n!n6npnn!n而limn!1npn!n=exp(limn!11nnXi=1lnin)=expZ10lnxdx=1elimn!1npnn!n=limn!1npnlimn!1npn!n=1e所以由夹逼准则知所求极限为1eExample7:求极限:limn!1nsinnn2+1+sin2nn2+2+


+sinn2+n!.Solution由于1n+1nXi=1sinin6nXi=1sininn+in61nnXi=1sinin而limn!11n+1nXi=1sinin=limn!1n(n+1)limn!1nnXi=1sinin=1Z0sinxdx=2by汤第2页,共9页利用定积分定义求极限2进阶题limn!11nnXi=1sinin=1limn!1nnXi=1sinin=1Z0sinxdx=2所以由夹逼准则知所求极限为2Example8:求极限:limn!1n2Xk=1nn2+k2.Solution法1.一方面limn!1n2Xk=1nn2+k2=limn!1n2Xk=1Zkk1nn2+k2dx6limn!1n2Xk=1Zkk1nn2+x2dx=Zn20nn2+x2dx=2另一方面limn!1n2Xk=1nn2+k2=limn!1n2Xk=1Zk+1knn2+k2dxlimn!1n2Xk=1Zk+1knn2+x2dx=Zn2+11nn2+x2dx=2故由夹逼准则知limn!1n2Xk=1nn2+k2=2法2.设Sn=limn!1n2Xk=1nn2+k2=n2Xk=111+kn
2
1n因Zk+1nkndx1+x211+kn
2
1nZknk1ndx1+x2则Z1nn2+1ndx1+x2SnZn0dx1+x2当n!1时，该不等式左右两端的极限都趋于Z+10dx1+x2=2由夹逼准则可知原极限为22进阶题Example1:求100Xn=1n12的整数部分by汤第3页,共9页利用定积分定义求极限2进阶题.Solution一方面100Xn=1n12=1+100Xn=2n12=1+100Xn=2Znn1n12dx1+100Xn=2Znn1x12dx=1+Z1001x12dx=19或者100Xn=1n12Z10111qx12dx=2p100:5p218:636另一方面100Xn=1n12=100Xn=1Zn+1nn12dx100Xn=1Zn+1nx12dx=Z1011x12dx=2p101118:1因此100Xn=1n12的整数部分为18.Example2:求极限:limn!1p1
2n2+1+p2
3n2+2+


+pn
(n+1)n2+n!.Solution由于in2+n6pi
(i+1)n2+i6i+1n2+1而limn!1nXi=1in2+n=limn!1n2n2+nlimn!11nnXi=1in=1Z10xdx=12limn!1nXi=1i+1n2+1=limn!1nXi=1in2+1+limn!11n2+1=limn!1n2n2+1limn!11nnXi=1in=1Z10xdx=12故由夹逼准则知limn!1p1
2n2+1+p2
3n2+2+


+pn
(n+1)n2+n!=12Example3:求极限:I=limn!11p+3p+


+(2n1)pnp+1.Solution考虑f(x)=xp(x2[0;2]).将[0;2]n等分,分点为2in,(i=1;2;


;n),小区间长度为∆xi=2n(i=1;2;


;n),取i=2i1n(i=1;2;


;n),=maxf∆xig=2n,by汤第4页,共9页利用定积分定义求极限2进阶题故I=12limn!12nnXk=12k1np=12lim!0nXi=1(i)p∆xi=12Z20xpdx=2pp+1Example4:求极限:limn!11nnXi=1sini12n!.Solution法1.考虑f(x)=sin(x)(x2[0;1]).将[0;1]n等分,分点为in,(i=1;2;


;n),小区间长度为∆xi=1n(i=1;2;


;n),取i=i12n(i=1;2;


;n),=maxf∆xig=1n,故limn!11nnXi=1sini12n!=lim!0nXi=1sin(i)∆xi=Z10sin(x)dx=2法2.考虑f(x)=sinx(x2[0;]).将[0;]n等分,分点为in,(i=1;2;


;n),小区间长度为∆xi=n(i=1;2;


;n),取i=i12n(i=1;2;


;n),=maxf∆xig=n,故limn!11nnXi=1sini12n!=1limn!1nnXi=1sini12n!=1lim!0nXi=1sin(i)∆xi=1Z0sinxdx=2Example5:求极限:limn!11+3+


+(2n+1)+12+4+


+(2n)+1(;¤1).SolutionI=limn!11+3+


+(2n+1)+12+4+


+(2n)+1(;¤1)=2limn!18:2n1n+3n+


+2n+1n#9=;+18:2n2n+4n+


+2nn#9=;+1=2Z20xdx+1Z20xdx+1=2(1+1x+120)+1(1+1x+120)+1=2(+1)+1(+1)+1by汤第5页,共9页利用定积分定义求极限3高级题Example6:求极限:I=limn!1nXk=1k(n+k)(n+k+1).SolutionI=limn!1nXk=1kn+kkn+k+1=limn!11n+11n+2+2n+22n+3


+nn+nnn+n+1=limn!1nXk=1kn+k!nn+n+1!=limn!11nnXk=111+kn12=Z1011+xdx12=ln212Example7:求极限:limn!1b1n1
n1Xi=0binsinb2i+12n(b1):.Solution考虑sinx在[1;b]上按以下划分1=b0nb1nb2n


bnn=b所做的积分和,其中∆i=bi+1nbin为小区间bin;bi+1n的长度,最大区间长度=max06i6n1f∆ig6b(b1n1)!0;又i=b2i+12n为小区间两端点的比例中项,因此原式=limn!1n1Xi=0sini‚…„ƒb2i+12n∆xi‚…„ƒbi+1nbin
=Zb1sinxdx=cos1cosb3高级题Example1:求积分:Z+10xx3dxby汤第6页,共9页利用定积分定义求极限3高级题.Solution当n6x6(n+1)时Z+10xx3dx=1Xi=0Z(i+1)ixx3dx=1Xi=0Z(i+1)iix3dx=1Xi=0i221i21(i+1)2=1221Xi=01k1k+1+1(k+1)2=112其中:limn!11+122+132+


+1n2=26Example2:设an=cosnpncos2npn


cosnnpn,求limn!1an+Proof:取对数,我们有lnan=lncosnpncos2npn


cosnnpn=nXk=1lncosknpn=nXk=1ln1+cosknpn1从而可得limn!1lnan=lnlimn!1an=lnlimn!1nXk=1k222n3k4412n6+o1n2=lnlimn!1nXk=1k222n3+nXk=1k4412n6+o1n2!=22Z10x2dx+0=26于是limn!1an=e26Example3:求极限:limn!1nn+1(11n+21n+


+n1n)n.Solution注意到11n+21n+


+n1n=k1nx1n‚…„ƒnXk=1Zkk1k1ndxnXk=1Zkk1x1ndx=Zn0x1ndx=n
n1n+1n+111n+21n+


+n1nZn+11x1ndx=n[(n+1)1n+11]n+1而limn!1nn+1n
n1n+1n+1n=limn!11+1nn=eby汤第7页,共9页利用定积分定义求极限3高级题limn!1nn+1n[(n+1)1n+11]n+1n=limn!1n(n+1)n(n+1)1n+11=(explimn!1n(n+1)1n+1!#)1=e由夹逼准则得limn!1nn+1(11n+21n+


+n1n)n=eExample4:求极限:limn!1241nnXk=1(lnk)21nnXk=1lnk!235.Solution(by欧阳)由于1nnXk=1ln2k1nnXk=1ln2kn=1nnXk=12lnlnkln2n
1nnXk=1lnk!21nnXk=1lnkn!2=1nnXk=12lnlnkln2n
于是limn!1241nnXk=1(lnk)21nnXk=1lnk!235=limn!1241nnXk=1ln2kn1nnXk=1lnkn!235=Z10ln2xdxZ10lnxdx2=1或者可以原式=limn!1241nnXk=1lnkn+lnn21nnXk=1lnkn+lnn!235将平方