第一章总练习题
1.求解下列不等式:5x?8（）1?2.3|5x?8|142解?2.|5x?8|?6,5x?8?6或5x?8??6,x?或x?.3552(2)x?3?3,52解?3?x?3?3,0?x?15.5(3)|x?1|?|x?2|1解(x?1)2?(x?2)2,2x?1??4x?4,x?.22.设y?2x?|2?x|,试将x表示成y的函数.1解当x?2时,y?x?2,y?4,x?y?2;当x?2时,y?3x?2,y?4,x?(y?2).3?y?2,y?4?x??1(y?2),y?4.??313.求出满足不等式1?x?1?x的全部x.2解x??1. 21?x?x?2,4(1?x)?x2?4x?4,x2?0.x??1,x?0.4.用数学归纳法证明下列等式:123nn?2(1)?2?3??n?2?n.222221?21证当n?1时,2-1?,等式成立.设等式对于n成立,则22123n?1123nn?1?2?3??n?1??2?3??n?n?1222222222n?2n?12n?4?(n?1)(n?1)?3?2?n?n?1?2??2?,n?1n?12222即等式对于n?1也成立.故等式对于任意正整数皆成立.(2)1?2x?3x?2?nxn?11?(n?1)xn?nxn?1?(x?1).(1?x)21?(1?1)xn?1x1?1(1?x)2证当n?1时,??1,等式成立.22(1?x)(1?x)设等式对于n成立,则1?2x?3x?2
?nxn?11?(n?1)xn?nxn?1?(n?1)x??(n?1)xn2(1?x)n1?(n?1)xn?nxn?1?(1?x)2(n?1)xn?(1?x)21?(n?1)xn?nxn?1?(1?2x?x2)(n?1)xn?(1?x)21?(n?1)xn?nxn?1?(xn?2xn?1?xn?2)(n?1)?(1?x)21?(n?1)xn?nxn?1?(xn?2xn?1?xn?2)(n?1)?(1?x)21?(n?2)xn?1?(n?1)xn?2?,2(1?x)即等式对于n?1成立.由归纳原理,等式对于所有正整数都成立.|2?x|?|x|?25.设f(x)?x(1)求f(?4),f(?1),f(?2),f(2)的值;(2)将f(x)表成分段函数;(3)当x?0时f(x)是否有极限:(4)当x??2时是否有极限?解(1)f(?4)?2?4?21?1?2?2?24?2?2??1,f(?1)??2,f(?2)??2,f(2)??0.?4?1?22??4/x,x??2;?(2)f(x)??2,?2?x?0;?0,x?0.?(3)无因为.limf(x)?2,limf(x)?0?limf(x).x?0?x?0?x?0?(4)有.limf(x)?lim(?4/x)?2,limf(x)?lim2?2?limf(x),limf(x)?2.x??2?x??2?x??2?x??2?x??2?x??26.设f(x)?[x2?14],即f(x)是不超过x2?14的最大整数.?3?(1)求f(0),f??,f(2)的值;?2?(2)f(x)在x?0处是否连续?(3)f(x)在x?2处是否连续?1??3??9??解(1)f(0)?[?14]??14,f?????14????6????7.f(2)?[?12]??12.4??2??4??(2)连续因为.limf(x)?lim[y?14]??14?f(0).x?0y?0?(3)不连续因为.limf(x)??12,limf(x)??11.x?2?x?2?7.设两常数a,b满足0?a?b,对一切自然数n,证明:bn?1?an?1bn?1?an?1nn(1)?(n?1)b;(2)(n?1)a?.b?ab?a
bn?1?an?1(b?a)(bn?bn?1a?证?b?ab?abn?1?an?1类似有?(n?1)an.b?an?an)?bn?bn?1b??bn?(n?1)bn,?1??1?8.对n?1,2,3,,令an??1??,bn??1??.?n??n?证明:序列{an}单调上升,而序列{bn}单调下降,并且.an?bn.证令a=1??1??1???n?n?1n?111,b?1?,则由7题中的不等式,n?1nn?11????1???n?1?11?nn?11????1???n?1?n?1??(n?1)?1??,?n?1?1??(n?1)?1???n?n(n?1)n?1nn?1??1???n??1??1???n?n?1n?1n?11??1?1???1????1???n?n?n?1?n?1,1??1??1??1??????n??n?1?n.n?1n?11??1??1??1?n???1??n?n?1????(n?1)?1???11?n?1??nn?11?1??1?(n?1)?1????1??n?1n(n?1)???n?1?1?1??1?????1???n?1?n?n?nnn?1nn?11????1???n?1?n?1n?11????1???n?1?n?11??11??1??1??1???????1??n?1??n??n?1??n2.111??我们证明?1???1??.nn?1?n?1?1121??1??1??nn?1n?1(n?1)211??.最后不等式显然成立.2n(n?1)(n?1)?1??1?当n??时,?1???e,?1???n??n?9.求极限
nn?1
?1??1??e,故?1???e??1???n??n?nn?1.1??1??1??1??lim?1?2??1?2??1?2??1?2?n???2??3??4??n?1??1??1??1??解?1?2??1?2??1?2??1?2??2??3??4??n?132435nn?11n?11???(n??).223344nnn22
nx10.作函数f(x)?lim2(a?0)的图形.n??nx?a?0, x?0;nx解f(x)?lim2??n??nx?a?1/x,x?0.
11.在?关于有界函数的定义下,证明函数f(x)在区间[a,b]上为有界函数的充要条件为存在一个正的常数M使得|f(x)|?M,?x?[a,b].证设存在常数M1,N使得M1?f(x)?N,?x?[a,b],取M?max{|M1|,|N|}?1,则有|f(x)|?M,?x?[a,b].反之,若存在一个正的常数M使得|f(x)|?M,?x?[a,b],则?M?f(x)?M,?x?[a,b].12.证明:若函数y?f(x)及y?g(x)在[a,b]上均为有界函数,则f(x)?g(x)及f(x)g(x)也都是[a,b]上的有界函数.证存在M1,M2,|f(x)|?M1,|g(x)|?M2,?x?[a,b].|f(x)?g(x)|?|f(x)|?|g(x)|?M1?M2,|f(x)g(x)|?|f(x)||g(x)|?M1M2,?x?[a,b].13.证明:f(x)?1?cos在x?0的任一邻域内都是无界的,但当x?0时f(x)不是无穷大量.xx11证任取一个邻域(??,?),??0和M?0,取正整数n,满足??和n?M,则f()?n?M,nn1故f(x)在(??,?)无界.但是xn??0,f(xn)?(2n?1/2)cos(2n?1/2)??0??,2n?1/2故当x?0时f(x)不是无穷大量.14.证明limn(x?1)?lnx(x?0).n??11lnx证令x?1?yn,则lnx?ln(1?y),n?.limyn?limxn?1?0.n??nln(1?y)n??1n1nln(1?y)注意到lim?limln(1?y)y?lnlim(1?y)y?lne?1,y?0y?0y?0y我们有n(x?1)?1n11ynlnx?lnx(n??).ln(1?yn)15.设f(x)及g(x)在实轴上有定义且连续.证明:若f(x)与g(x)在有理数集合处处相等,则它们在整个实轴上处处相等.证任取一个无理数x0,取有理数序列xn?x0,f(x0)?limf(xn)?limg(xn)?g(x0).n??n??16.证明lim1?cosx1?.2x?0x22sin2x22??1?cosx2siny1siny1212?lim证lim?lim?lim?1?.?y?0?222x?0x?0y?0xx4y2?y?22ln(1?y)ex?a?ex17.证明:(1)lim?1;(2)lim?ea.y?0x?0yxln(1?y)证(1)lim?limln(1?y)y?lnlim(1?y)y?lne?1.y?0y?0y?0yex?a?eaea(ex?1)ex?1ay1a(2)lim?lim?elim?elim?eax?0x?0x?0y?0ln(1?y)ln(1?y)xxxlimy?0y1?ea?ea.118.设y?f(x)在a点附近有定义且有极限limf(x)?0,又设y?g(x)在a点附近有x?a11定义,且是有界函数.证明limf(x)g(x)?0.x?a证设|g(x)|?M,0?|x?a|??0.对于任意??0,存在?1?0,使得当0?|x?a|??1时|f(x)|??/M.令??min{?1,?0},则0?|x?a|??时,|f(x)g(x)|?|f(x)||g(x)|??MM??,故limf(x)g(x)?0.x?a19.设y?f(x)在(??,??)中连续,又设c为正的常数,定义g(x)如下?f(x) 当|f(x)|?c?g(x)??c 当f(x)?c??c 当f(x)??c?试画出g(x)的略图,并证明g(x)在(??,??)上连续.
北大版高等数学第一章 函数及极限答案 第一章总练习题
