《經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)》形考作業(yè)一

1、經(jīng)濟(jì)數(shù)學(xué)基礎(chǔ)形成性考核冊(cè)及參考答案(一)填空題1.limx0x-sinx.答案:2.設(shè)f(x)=x2+1,lk,x=0,在x=0處連續(xù)，則x=0.答案：13.曲線y=Jx+1在(1,2)的切線方程是.答案:24.設(shè)函數(shù)f(x+1)=x十2x+5,則f'(x)=.答案：2x5.設(shè)f(x)=xsinx,則f.答案:(二)單項(xiàng)選擇題1.當(dāng)XT+如時(shí)，下列變量為無(wú)窮小量的是(sinx答案:A.ln(1+x)B.12C.exD.2.下列極限計(jì)算正確的是()答案：B|xA.lim一x)0x二1C.limxsin1=1xTX3.設(shè)y=lg2x,則dy=B.lim一二1x-0'xsinxD.l

2、im=1x立x).答案：BA.dx2x4.若函數(shù)A.函數(shù)1B.一dxxln10f(x)在點(diǎn)xo處可導(dǎo)，則f(x)在點(diǎn)xo處有定義cln10.C.dxx)是錯(cuò)誤的.答案:B.函數(shù)f(x)在點(diǎn)Xo處連續(xù)C.limxxof(x)=A，但A#f(x0)D,函數(shù)f(x)在點(diǎn)xo處可微5.若f(1)=x,則f(x)=(.答案：BA.1.x12x計(jì)算極限1B-xC.1D.-xX2-3x2x27=lxm1(x1)(x2)(x7)(x1)lim"Ix1解：ximx2-5x6x2-6x8=lim(x-3)(x-2)x，2(x-4)(x-2)=limCx屹x4解：(3)解:(4)22x-3x5lim2x1

3、3x2x4=lim2x2x>:3xsin3x-3x解:(5).sin3x3xlim=limx0sin5xx0sin5x5X5x解:(6)x2 -4lim 二 limx 2sin(x -2) x n(x - 2)(x 2)x - 2 二 lim lim (x 2) = 4sin(x -2) x 祥 sin(x -2) x)22.設(shè)函數(shù)f(X)=.1 .xsin - b,xa,X : 0sin xx問(wèn)：(1)當(dāng)a, b為何值時(shí)，f (x)在x=0處有極限存在(2)當(dāng)a,b為何值時(shí)，f (x)在x = 0處連續(xù).解： lim f (x) = lim sin =1;x-0 'x_0 &#

4、39; x1,、,lim f(x) = lim (xsin b)=b;f(0)=ax Q x0 -x二(1)當(dāng)b=1, a任意時(shí)，f (x)在x=0處有極限存在;(2)當(dāng)a=b=1時(shí)，f (x)在x = 0處連續(xù)。3 .計(jì)算下列函數(shù)的導(dǎo)數(shù)或微分:(1)2 x2y=x +2 + log 2 x - 2，求y解:v1y'=2x 2Xln2 xln2(2)ax b,求ycx d解:(ax b) (cx d) (ax b)(cx d) a(cx d) (ax b)c2(cx d)2(cx d)ad -cb2(cx d)(3)1.3x -5解:.1y'= 13x-5) 21-(3x-5)

5、3_3(3x-5) = -(3x-5) 2-'32 V (3x-5)3(4)y = . x -xeX ,求 y解:11Vy'=T(xexx(ex)(x1)ex2,x2x(5)ax二esinbx,求dy解:(6)xXxX,求dy解:y=(ex)(x2)1=ex(1)x3x：2iex-2xy二cos、,x2x,求dy解：；y = (-sin . x) ( . x) -2x ln 2二(-sin、x)21-MV"/2(8)y =sinn x +lnx，求 y'解:n1xcosx 一 x(9)y =e+ cos2x，求 y解:(10) y1 sin =2 x1 3 x2 - 2x x解:111sin -y(2 xx2x6- . 2): 2i sin-x,c , . 1、1ln 2 (sin -) x x2321 sin-2 x1-1ln 2. cosx35二1一6x 2 x 66n11y=nsinx(sinx)-=nsinx4.下列各方程中y是x的隱函數(shù)，試求y或dy(1) x2+y2-xy+3x=1,求dy解：兩邊對(duì)x求導(dǎo)：2x+2yy_(y+xy)+3=02yy -xy =y2x3貝U y2y - x(2) sin(x+y)+exy=4x,求y解：兩邊對(duì)x求導(dǎo)：cos(x+y)1十y)exy(y xy)4-cos(xy)-yexyc