答案：本科高数作业卷(上册)new

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1、答案:本科高等数学作业卷(一)一、填空题x−xe−e.1极限lim是否存在______.x−xx→∞e+exx解lime=+∞lime=0x→+∞x→−∞x−x−2xx−x2xe−e1−ee−e−1+e∴lim=lim=1lim=lim=−1x→+∞ex+e−xx→+∞1+e−2xx→−∞ex+e−xx→−∞1+e2xx−xe−e所以lim不存在.x−xx→∞e+e2.2limx(x+100+x)=______.x→−∞2100x100解limx(x+100+x)＝lim=lim=−50x→−∞x→−∞2x→−∞100(x+100−x)−1+−12xx⎛x+2

2、a⎞.3设lim⎜⎟=8，则a=____.x→∞⎝x−a⎠3ax−ax−a3a⋅3a+a⎡3a⎤3aa3a解左边=lim1(+)3a=lim1(+)3a⎥⋅lim1(+)=e⎢x→∞x−a⎣x→∞x−a⎦x→∞x−a3a∵e=8∴a=ln2f(x)−3.4设f(x)在x=2连续，且lim存在，则f)2(=______.]x→2x−2f(x)−3解由lim存在知：limf(x)=3，x→2x−2x→2又根据f(x)在x=2连续得：f)2(=limf(x)=.3x→2⎧⎪2xxe,x<0⎪.5设f(x)=⎨b,x=0,当a=___,b=___时，f(x)在(−∞,

3、+∞)连续.⎪ln(1+2x)⎪+a,x>0⎩x2xln(1+2x)2x解limf(x)=limxe=,0limf(x)=lim[+a]=lim+a=2+a,−−+++x→0x→0x→0x→0xx→0x根据连续的定义limf(x)=limf(x)=f)0(得：−+x→0x→0⎧2+a=0⎧a=－2⎨,即⎨.⎩b=0⎩b=01二、选择题21x−1.1当x→1时，函数ex−1的极限x−1(A)等于2()B等于0(C)为∞(D)不存在但不为∞11解limf(x)=lime2x−1=+∞limf(x)=lime2x−1=0x→1+x→1+x→1−x→1−所以应选(D)

4、.2x.2已知lim(−ax−b)=0，其中a,b是常数，则x→∞x+1(A)a=,1b=1()Ba=−,1b=1(C)a=,1b=−1(D)a=−,1b=−122x1+(x−)11解lim(−ax−b)=lim[−ax−b]=lim+lim[(1−a)x+(−1−b)]x→∞x+1x→∞x+1x→∞x+1x→∞=lim1(−a)−1(+b)=0x→∞⎧1−a=0⎧a=1所以有⎨，解得⎨.故应选(C).⎩1+b=0⎩b=－1⎧x−,10

5、(C)limf(x)不存在(D)limf(x)不存在x→1+x→1解limf(x)=lim(x−)1=,0limf(x)=lim2(−x)=1故应选(D).−−++x→1x→1x→1x→1atanx+b1(−cosx)22.4设lim=2，其中a+c≠0，则必有2x→0−xcln(1−2x)+d1(−e)(A)b=4d()Bb=−4d(C)a=4c(D)a=−4ctanx1−cosxtanx1−cosxa⋅+b⋅a⋅lim+b⋅lim解左边=limxx=x→0xx→0x22x→0−x−xln(1−2x)1−eln(1−2x)1−ec⋅+d⋅c⋅lim+d⋅li

6、mxxx→0xx→0x12xx2a⋅lim+b⋅limx→0xx→0xa==−2−2xx2cc⋅lim+d⋅limx→0xx→0xa由−=2得a=−4c故应选(D)2c2三、计算、证明题12+exsinx.1求lim(+).x→04x1+ex143－－2+exsinxe2x+exsinx解limf(x)=lim(+)=lim(+)++4+4x→0x→0xx→0－x1+exex+1=0+1=112+exsinxlimf(x)=lim(−)=2−1=1−−4x→0x→0x1+ex故limf(x)=1x→021sinx+xsinx.2limx→01(+cosx)ln

7、(1+x)2121sinx+xsinsinx+xsinx1x解lim=limlimx→01(+cosx)ln(1+x)x→01+cosxx→0x1sinx11=[lim+limxsin]=.2x→0xx→0x2⎛12n⎞.3lim⎜++"+⎟n→∞⎝n2+n+1n2+n+2n2+n+n⎠1+2+"+n12n1+2+"+n解≤++"+≤22222n+n+nn+n+1n+n+2n+n+nn+n+1n(n+)11+2+"+n21而lim=lim=22n→∞n+n+nn→∞n+n+n2n(n+)11+2+"+n21lim=lim=n→∞n2+n+1n→∞n2+n+12

8、1所以由夹逼定理得：原式=2nnln(