利用定积分定义求极限(by汤).pdf

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1、利用定积分定义求极限1入门题同济7的p226ZbXnf(x)dx=I=limf(i)∆xi;xi166xia!0i=1做题时的公式Z1Xn1if(x)=limf()n!1nn0i=1ZbXnbaif(x)dx=limf(a+(ba))n!1nna„ƒ‚…i=1∆x1入门题111Example1:求极限I=lim++


+n!1n+1n+22n.Solution!1111I=lim++


+12nn!1n1+1+1+nnnXnZ1111=lim=dx=ln2n!1n1+i01+xi=1n1n!nExample2:求极限:limn!1nn.Solution1

2、n!n1n!1n!lim=limexpln=explimlnn!1nnn!1nnnn!1nnn112n=explimln+ln+


+lnn!1nnnnXnZ11i=explimln=explnxdxn!1nn0i=1hiZ111=expxlnxdx=00eq1nExample3:求极限:I=limn(n+1)(n+2)


(2n1)n!1n.SolutionXn1!Z11i4I=explimln1+=expln(1+x)dx=n!1nn0ei=11111Example4:求极限:I=limp+p+p+


+pn!112+n222+n232+n2

3、n2+n2by汤第1页,共9页利用定积分定义求极限1入门题.SolutionXn11Xn1I=limp=limqn!1i2+n2n!1n(i)2+1i=1i=1nZ11hpi1=pdx=ln(x+x2+1)0x2+10p=ln(1+2)12nExample5:求极限:lim++


+n!112+n222+n2n2+n2.Solution12nI=lim++


+n!112+n222+n2n2+n2Xni1Xnin=nlim!1i2+n2=nlim!1ni
2i=1i=1n+1Z1Z1x112=dx=d(1+x)01+x2201+x21112=ln(1+x)=ln22

4、02pn1+2!+3!+


+n!Example6:求极限:limn!1n.Solution由于pppnnnn!1+2!+3!+


+n!nn!66nnn而p()Znn!1Xni11lim=explimln=explnxdx=n!1nn!1nn0ei=1ppnnnn!pn!1nlim=limnlim=n!1nn!1n!1ne1所以由夹逼准则知所求极限为e!2sinsinsinnnExample7:求极限:limn++


+n!1n2+1n2+2n2+n.Solution由于1XniXnsini1Xninsin66sinn+1nn+inni=1i=1ni=1

5、而1XninXni1Z2limsin=limlimsin=sinxdx=n!1n+1nn!1(n+1)n!1nn0i=1i=1by汤第2页,共9页利用定积分定义求极限2进阶题1Xni1Xni1Z2limsin=limsin=sinxdx=n!1nnn!1nn0i=1i=12所以由夹逼准则知所求极限为n2XnExample8:求极限:limn!1n2+k2k=1.Solution法1.一方面Xn2Xn2Zknnlim=limdxn!1n2+k2n!1k1n2+k2k=1k=1Xn2ZkZn2nn6limdx=dx=n!1k1n2+x20n2+x22

6、k=1另一方面Xn2Xn2Zk+1nnlim=limdxn!1n2+k2n!1kn2+k2k=1k=1Xn2Zk+1Zn2+1nn>limdx=dx=n!1kn2+x21n2+x22k=1故由夹逼准则知n2Xnlim=n!1n2+k22k=1法2.设Xn2Xn2n11Sn=nlim!1n2+k2=k
2
nk=1k=11+n因Zk+1Zkndx11ndx2<k
2
<2k1+x1+nk11+xnnn则Z1Znndxdx

7、le1:求n2的整数部分n=1by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X100X100X100Zn111n2=1+n2=1+n2dxn=1n=2n=2n1X100ZnZ10011<1+x2dx=1+x2dx=19n=2n11或者X100Z101pp11n2