例3 设a?0,x1?a,x2?a?a,…, xn?a?a???a.求limxn.
n???解 容易证明{xn}单调递增.现证对任意自然数n,xn?xn?a?1.则xn?1?a?xn?a?1.当n?1时显然成立.归纳假设
a?a?1?a?2a?1?a?xna?1.
由单调有界定理, {xn}有极限.设limxn?x.对xn?1?n???两边取极限得x?a?x.
解得x?
1?1?4a2.由于x?0,故得limxn?n???1?1?4a2.
例4 设x1?4,当n?2时, xn?1xn?12?xn?12.求limxn.
n???解 显然xn?0.由于xn?1?xn?xn?2?2xn2xn2?2?xn2xn2与xn?1xn?1?xn?12?21xn?1?xn?12?2
所以xn?1?xn?0,即{xn}单调递减且有下界.故{xn}极限存在,令limxn?x.
n???由递推关系式得x
?1x?x2.解得x?2,即limxn?n???2.
例5 设x1?0,且对任意自然数n,xn?12?xnxn?3a3x2n?2?其中a?0?a.求limxn.
n???解 由于
xn?1xn?xn?3a3x2n,
?a2xn?1xn?1?xn?3a?3xn?a?xn?a222,与xn?1?a2?xnxn?3a?2?2?a?3xn?a?222(3xn?a)2
?x?x?n2n?3a??a3xn?a(3x2n?2???x?xn2n?3a??a3xn?a?2?a)2n2??
?x?n?a??x32n?2a?3(3xn?a)??x?a?3(3xn?a)22
故xn2?a与xn2?1?a同号.因此当x12?a时有xn2?a(n?1,2,?),此时{xn}递增有上界a;当x12?a时有xn2?a(n?1,2,?),此时{xn}递减有下界a. 所以{xn}收敛,设limxn?x.则xn????xx3x?22?3a?a?.因为x?0,解得x?a,即limxn?n???a.
例6 设xn?1?12?13???1n?lnn,证明{xn}收敛.
11证明 由xn的定义, xn?1?xnn?1??1??????ln(n?1)?lnn??ln(1?).由于??1???n?1n?1nn??????1单
1?调递减趋于e,故??1??n??n?1?e.取对数得?n?1?ln(1?1n)?1,ln(1?1n)?1n?1.所以这证明了
{xn}单调递减.
n?1????又由于??1????n??????单调递增趋于e,可得不等式
1n1n?ln(1?1n).
因此1?12?13????(ln2?ln1)?(ln3?ln2)????ln?n?1??lnn??ln?n?1??lnn.
所以xn?0(n?1,2,?),由单调有界定理,{xn}收敛.设limxn?C,这里C称为Euler常数.
n???可以证明0?C?1(C?0.5775216?).
例7 设x1?1,x2?12,
xn?1?11?xn.求limxn.
n???解 若极限存在,设为A,则A?因xn?0(n?1,2,?),A若xn?A,则xn?1?11?xn11?A,A2?A?1?0,A??1?211?xn5?0.618??????1.618?.
?0.618?.若xn?A,则xn?1?A??11?A?A;
?0.618??11?A.即xn在A的左右来回跳动,而x1?1
知:x1,x3,x5,?,x2n?1,??A,x2,x4,x6,?,x2n,??A (1).
若{xn}收敛于A,则{x2n},{x2n?1}也收敛于A.猜想:是否{x2n}在A左端单调递增到
A,{x2n?1}在A右端单调递减到A.下面来考xn?2?xn察的符号.
xn?2?xn?11?11?xn211?xn?1?xn
?xn
??1?xn?xn2?xn
?(1.618??xn)(0.618??xn)??0,???0,2?xn?若xn?0.618??A,若xn?0.618??A. (2).
式(1),(2)表明{x2n}以A为上界, {x2n?1}以A为下界. 因此二子列收敛.记limx2n??,limx2n?1??.
n???n???在式x2n?1?11?x2n及x2n?11?x2n?1n???中令n???,有??11??,??11??.
所以????A?0.618?.既然limx2n?limx2n?1?A,故limxn?A?0.618?.
n???n???
例8 证明序列2,2?12,2?12?12,?收敛,并求其极限.
解 以序列特征可以看出,相邻两项的关系是xn?1满足方程A?2?1A?2?1xn (1).因此,设{xn}收敛,则极限A2??.又xn?0,所以A2??1?2.令xn?A??n?1?n (2).(2)代入
(1), ?n?1?2?1?1??nn2?? (3).则将满足(1)的序列{xn}?A的问题,转化为满足(3)的序列
12{?n}?0的问题.事实上, ?1?x1?A?1?,即limxn?lim(1?n???2,?1?.由(3)利用数学归纳法,易证
?n?12nn???2??n)?1?2.
§1.4 求函数极限的几种方法
一、利用函数的连续性求极限
定理 (复合函数求极限定理) 设函数f(y)在y?b连续,函数y?g(x)有性质limg(x)?b,则
x?alimf[g(x)]?f[limg(x)]?bx?a.
u(x)v?x?x?a推论 设limu(x)?A?0,limv(x)?B,则limx?a?AB.
x?ax?a证明 由复合函数求极限定理,
limu(x)x?av?x? ?limex?av?x?lnu?x??ex?alimv?x?lnu?x??eBlnA?AB
例1 求极限limx?0ax?1x.(a?0)?0
时y?0.解得xlnaln?1?y?y1解 令ax?1?y,则当x故limx?0?ln?1?y?lna.
ax?1x?limy?0ylnaln?1?y??limy?0?lna.
注 此例中取x 例
?1n?0,得数列极限, limn?na?1??0.(a?0)
n????2 求极限lim?n?????nna?2nb??,?a?0,b?0? ??n解 令由于
a?2nb?1?xnn,则xnnn?a?2nb?1?0(n???).
n???limxn??limn??n????a?2b?n?1??lim?n???2??na?1?nb?1??12?lna?lnb??lnab,
?所以lim?n?????na?2nb????n??lim(1?xn)?lim?1?xnn???n?????n?x1n???nxn?elnab?ab.
1sinx?例3 求极限lim???x?ax?a.?a?k??
sinx?sina1sina?sina?11sinx?解 lim???x?asina??x?a??sinx???lim?1???1??x?a?sina???x?a?sinx?sina?sin??lim??1??x?a?sina???1cosasinax?sina????x?a?.
由于limx?asinx?sinax?a?sinx??cosa,所以lim??x?asina??x?a?esina?ecota.
例4求极限limx?0xcosx.
?12cosx?1x解 注意到limx?0cosxx?1,我们有
1alimx?0xcosx?lim1?cosx?0??x?1????lim?1?cosx?0???x?11??1cosx?1????e.
2练习 求极限 (1)
?cosx?lim??x?0cos2x??x2?; (2) lim?2?3xxx?0??2?3xx22????x.
二、利用微分学方法(L’Hospital法则,Taylor公式)求极限 例5 求极限limx?0?1?x?xx1?e.
?u'?x????v'?x?lnu?x??v?x??u?x???解 由导数公式
ddx1ddxu(x)v?x??u(x)v?x?
得
1(1?x)x?(1?x)xx??1?x?ln?1?x?x2?1?x?
由L’Hospital法则得
limx?0?1?x?xx1?e?elimx?0x??1?x?ln?1?x?x2?1?x??elimx?0?ln?1?x?2x?3x2??e2.
例6 求极限lim?1?2?2?cotx?. x?0?x?解 利用L’Hospital法则与等价无穷小代换得
?1lim?2?cotx?0?x2sin?x??limx?0?2x?xxsin222cosx2x?limx?0sin2x?xx42cos2x(等价无穷小代换)
?limx?0sinx?xcosxxsinxx?sinx?xcosxx3(化简) ?2limx?0cosx?cosx?xsinx3x2(L’Hospital法则)
?23limx?0?23.
1例7求极限lim?a?x?0??x?bx?cx3????x
解 由指数函数的连续性与L’Hospital法则得
1?alim?x?0??x?bx?cx3????xxxx?a?b?cln?3?exp?limx?x?0??3???a?explim??x?0????xlna?baxxlnb?cxx?b?cxlnc???
?lna?lnb?lnc??exp???3??abc.其中expy表示指数函数ey.