1、利用定积分定义求极限1入门题同济7的p226Z baf (x)dx = I = lim!0nXi=1f ( i)xi; xi 1 6 6 xi做题时的公式Z10f (x) = limn!1 1nnXi=1f (in)Z baf (x)dx = limn!1 b anxnXi=1f (a + in(b a)1入门题Example1:求极限I = limn!11n + 1 +1n + 2 + +12n. SolutionI = limn!1 1n11 + 1n +11 + 2n + +11 + nn!= limn!1 1nnXi=111 + in =Z 1011 + x dx = ln2Examp
2、le2:求极限: limn!1n!nn1n. Solutionlimn!1n!nn1n= limn!1exp 1n lnn!nn= exp limn!1 1n lnn!nn=exp limn!1 1nln 1n + ln 2n + + ln nn=exp limn!1 1nnXi=1ln in = expZ 10lnx dx=exphx lnxi10Z 10dx= 1eExample3:求极限: I = limn!1 1n nqn(n + 1)(n + 2) (2n 1). SolutionI = explimn!1 1nn 1Xi=1ln1 + in!= expZ 10ln(1 + x)dx
3、= 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=Z 101px2 + 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
4、 + n2= limn!1nXi=1ii2 + n2 = limn!11nnXi=1in in2 + 1=Z 10x1 + x2 dx =12Z 1011 + x2 d(1 + x2)=12 ln(1 + x2)10= 12 ln2Example6:求极限: limn!1np1 + 2! + 3! + + n!n. Solution由于npn!n 6np1 + 2! + 3! + + n!n 6npn n!n而limn!1npn!n = exp(limn!1 1nnXi=1ln in)= expZ 10lnx dx= 1elimn!1npn n!n = limn!1npn limn!1npn!
5、n =1e所以由夹逼准则知所求极限为1eExample7:求极限: limn!1nsin nn2 + 1 +sin 2n n2 + 2 + +sin n2 + n!. Solution由于1n + 1nXi=1sin i n 6nXi=1sin i nn + in 61nnXi=1sin i n而limn!1 1n + 1nXi=1sin i n = limn!1 n(n + 1) limn!1 nnXi=1sin i n = 1 Z 0sinx dx = 2 by汤第2页,共9页利用定积分定义求极限2进阶题limn!11nnXi=1sin i n = 1 limn!1 nnXi=1sin i
6、 n = 1 Z 0sinx dx = 2 所以由夹逼准则知所求极限为2 Example8:求极限: limn!1n2Xk=1nn2 + k2. Solution法1.一方面limn!1n2Xk=1nn2 + k2 = limn!1n2Xk=1Z kk 1nn2 + k2dx6 limn!1n2Xk=1Z kk 1nn2 + x2dx =Z n20nn2 + x2dx =2另一方面limn!1n2Xk=1nn2 + k2 = limn!1n2Xk=1Z k+1knn2 + k2dx limn!1n2Xk=1Z k+1knn2 + x2dx =Z n2+11nn2 + x2dx =2故由夹逼准则
7、知limn!1n2Xk=1nn2 + k2 =2法2.设Sn = limn!1n2Xk=1nn2 + k2 =n2Xk=111 + kn 21n因Zk+1nkndx1 + x2 100Xn=1Z n+1nx 12 dx =Z 1011x 12 dx = 2p101 118:1因此100Xn=1n 12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 + n 6pi (i + 1)n2 + i 6i + 1n2 + 1而limn!1nXi=1in2 + n = limn!
8、1n2n2 + n limn!11nnXi=1in = 1 Z 10x dx = 12limn!1nXi=1i + 1n2 + 1 = limn!1nXi=1in2 + 1 + limn!11n2 + 1= limn!1 n2n2 + 1 limn!11nnXi=1in = 1 Z 10x dx = 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!1 1p + 3p + + (2n 1)pnp+1. Solution考虑f (x) = xp (x 2 0;2).将0;2 n
9、等分,分点为2in , (i = 1;2; ;n),小区间长度为xi = 2n (i = 1;2; ;n),取 i = 2i 1n (i = 1;2; ;n), = maxfxig = 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 = 12 limn!1 2nnXk=12k 1np= 12 lim!0nXi=1( i)pxi = 12Z 20xp dx = 2pp + 1Example4:求极限: limn!1 1nnXi=1sini 12n !. Solution法1.考虑f (x) = sin( x)(x 2 0;1).将0;1 n等分,分点为in, (i = 1;2; ;n
10、),小区间长度为xi = 1n (i = 1;2; ;n),取 i = i 12n (i = 1;2; ;n), = maxfxig =1n,故limn!1 1nnXi=1sini 12n != lim!0nXi=1sin( i )xi =Z 10sin( x)dx = 2 法2.考虑f (x) = sinx (x 2 0; ).将0; n等分,分点为i n , (i = 1;2; ;n),小区间长度为xi = n (i = 1;2; ;n),取 i = i 12n (i = 1;2; ;n), = maxfxig =n,故limn!1 1nnXi=1sini 12n != 1 limn!1
11、nnXi=1sini 12n != 1 lim!0nXi=1sin( i)xi = 1 Z 0sinx dx = 2 Example5: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 1):. Solution考虑sinx在1;b上按以下划分1 = b 0n x 1n nXk=1Z kk 1k 1n dx nXk=1Z kk 1x 1n dx =Z n0x 1n dx = n n
12、1n+1n + 111n + 21n + + n1n ln(n + 1) + + ln(n + n)n + 1 lnnln(n + 1)n + 1 +ln(n + 2)n + 12 + +ln(n + n)n + 1n lnn 6ln(n + 1) + + ln(n + n)n lnn且limn!1ln(n + 1) + + ln(n + n)n lnn= limn!1ln(n + 1) lnn + ln(n + 2) lnn + + ln(n + n) lnn + nlnnn lnn= limn!1 1nln1 + 1n+ ln1 + 2n+ + ln1 + nn=Z 10ln(1 + x)
13、dx = 2ln2 1limn!1ln(n + 1) + + ln(n + n)n + 1 lnn= limn!1ln(n + 1) lnn + ln(n + 2) lnn + + ln(n + n) lnn + nlnnn + 1 lnn= limn!11n + 1ln1 + 1n+ ln1 + 2n+ + ln1 + nnlnn1 + n= limn!1 nn + 1 limn!1 1nln1 + 1n+ ln1 + 2n+ + ln1 + nn+ limn!1 lnn1 + n=1 Z 10ln(1 + x)dx + 0 = 2ln2 1所以由夹逼准则知所求极限是2 by汤第9页,共9页
