《经济数学基础微积分第二章习题解答重点》由会员分享，可在线阅读，更多相关《经济数学基础微积分第二章习题解答重点（14页珍藏版）》请在金锄头文库上搜索。
1、l.(6)un = af, (a 为常数)lima = 0 a 1或*-1lima = 1a = 1、一69.(1)x+ 2x2-l解：即xt时为无穷大量 ln(l-x)解:l-XT+oc或-A-()时为无穷大量即X -QOIOCV T 1丄(3)亦解:丄-p时为无穷大量，即兀T 0 X1y/ X 2解:y/x-2 T0时为无穷大量,即X T 2+解： + 2arctaar-0lll为无穷大壮即 arctan x，贝iJxtyc2兀2 + 1(依+ 1尸解：分子的最高次项秒数大于分母的最高汝项次数，Hx0.即KTP时为无穷大量13 求下列极限：I2(1) lim := lim=2W妇+”n卄/
2、 j(2) lim( I +yfn)= lim Vh( J1+ 1) = oon-oon-oV H(或 *.* lim !t=- = 0 原式=oo)i V/-1 +VZ/川、narctaiv/(3) lim亍匚n y/n2n(4) lim(l+； +当.卄 33*=limn-bbfl 00兀T1-(3r，1(5) lim(3x2-5x) =3x22-5x2 = 14 x-233132 +口）=2 + 口飞P_ Ic J3x-1v x x2 n(7) lim=11血=0is 2x + 42 + x(9)一1 一 Jx + 1).t(8)lim3._3 = jjL = 0 Z2.J+1 2 +
3、1=lim = 4-00宀厶2-1+坂+1(x-3),2(2x + 1)828（吧（3一(12) lim3丘 +5F +477 + 2(d + x) -a:a2 +2.v+x? -a(I l)lim= limx-40rx-or= lim)= 2aX(讪口戸=面(茫-3)(茫+ 1)f X _ 9KM (Vx - 3)(Vx + 3)(14)lim(.t+VT7)= lim (x + FF)(-*匸+(1-怛A-HO.r2-x Vl-r +(l-r )3x3 + l-x3= =0X2-Xyl-X3 +(1-X3)3a +b =(“+/?)(/ -abb2)十J4宀l+2x(15)lim x(IA
4、x2 - I -2.v)-4+2八 yfx + 3-2一 (77一2)(7!75 + 2)(頁4 1)1+11()忸乔厂巳吧石7二1)(広+i)(E+2)(i7)iim(V774-/77i)= iim_=J_= =P ()fl I14.WW3x-l x v 0(l)/(x) = sinx r0在x = 0处.X解：/(00)= lim(3x-l) =-1x-0*/(0 + 0) = lim =Iw xIn.r/(Tx-lX1()Xry/2-X -yx =lim1 一 X ”(I 2 JC + yx)limKT广In x71lim Inx7-*=ln( lim(l + .r 1)x 1 Jx-r
5、2x+lx(3)/(.v)= .v2 - x+3 I x 2解：x = 1. lim(2x+l) = 3lim(W -x+3) = 1-1 + 3 = 3x-rx-rx = 2Jim(x2 -a + 3) = 4-2 + 3 = 5Iim(x3-I) = 8-I = 7jtt215 .设有函 W*(x) = je* x0Tm竺空-血沁 =2-1 = 1tfO xKfO x(7)lim(l-)Jx-#0 tan 2x!2，sinF =!2，v =x(10)1曲匕妙“a xsinx丄丄Indnx) “卅一弘小血-血*(1 1) lim(l sin x)x= lime=占 x (一“小 =e .vO
6、x0(I 4)lim(v-0=Iim(l )c = e i v-o 2r _ 7=lim2(i+-rX(12)lim(! + 2x)K =e ,v-0(I6)lim(l -) r = exxf sr(17) 1 im ln( 1 + x + x2 + x): -o r x + x2 +x5=limtor(18)lim(r + 2 +3rt +4*1) = lim4xl + ( 1) + (2f +(3) F =4XT6XTS444I 7求下列极限:(1) lini(cos.r)Hco%x x-OIncosx= limeig _ xtO ln(1+cosx1)lim(COS.V1 )或lim(c
7、os x)1-cosx = lim【1-(1 - cos x)Ycosx = e x-0x0(2)lim(sec2 x)cc*2xO= lim(l + tan2 x)lan c-01 -2sinxnjr*弋 sin(x-)6sinr1 -2cosrsin 2sin/cossinr|iml-coS/-V3sin/=|im = 皿sin tz sin tlimsinx-tanx ：2 Sin AT= liinsinx(cosx-l)=iim)=()5 oosxsinx2 f cosx“、 sinx3 tanx吧1-cos?= lim - = 2TO |4-x7sin(2x)lan( x)(7) l
8、im t anx) t an( - x) = limsin(-2x) 厶 sin(2x)(-x) = lim4-2.v2/oxl. In Ji +5x In(l+5x)(8) lim= limz x 3 2x= lim zo 2x=5=2(a) sin 2xcos2x(7)lim tan(2x)tan-x) = lim 2兰42兰44吟X)(手-x)= lim=lim * . Ttxjrr4 sin(-2x)4(- - 2x)(p) 1=lim =x 兀o4 2(- -X)2221 J(9)limsin(l- )n = sinlim(l-)n =sine 22”/its 2n/inx r ln
9、(14-.r)-lnxln(l+)(10)limL=|imx_IlfWv11111人/IfHOX1-x191 -=lim ln(l + -)r = Ini = o n-wo%或=lim A = 0打一2兀.解:lim /(x)= lim /(x) = 2h/(-1)XT(-I)AX = 2是第一类可去间断*f () = 1+X0r./(-i)=z则为连续(2) x = 0第二类无穷间断点(3) x = 0第一类跳跃间断点(4) x = 0第一类可去间断点兀=1第二类无穷间断点(5U = 0第一类跳跃间断点(6)x = 0第一类可去间断点20讨论卜列西数的连续性0xlarctanv(l)/(x)
10、 =2-x(x- l)sin.r-1解：x = lim em = 1, lim(2-x) = 2 = f(0)右连续 kt(t x“ax-l)sinx-l = -l = /(l)右连续A=0,X=lib均不连续，但各门曲J内全为初等函数 连续区间为-1.0) 50,1) 51 ,+8)连续(-8.+00)(4)(一82)5 2.+OO)(3X-L0).(0.i).(l,l)22妣做収域)俅檄a和皿砌各门1 n sin xx 0X解：Df = (-oo.+oo)lim sin x = 1, lim(xsin +b) = b./(0) = a - 1 xx二 a = 2.b = 若想在赵义域连续，则/(0-0) = /(0) = /(0+0)23证明方程 ?X 1在(1,2内至少有一实根证明：设(工心 3x 1 显然它在1,2上连续f(l 3(U(210根零定，据值理在3 (1,2中，方X JX 1至有实。程少一根2 2 2(13) Hind )v-2= lim(l-)l(l-) = 2ATfSXXfSXX
