经济数学基础作业

导读:经济数学基础形成性考核册及参考答案,作业(一),经济数学基础形成性考核册及参考答案作业(一)(一)填空题x?sinx?___________________.答案:0x?0xx?sinxsinxsinx?lim(1?)?lim1?lim?1?1?0分析:limx?0x?0x?0x?0xxx1.lim?x2?1,x?02.设f(x)??,在x?0处连续,则k?________.答案:1?k,x?

经济数学基础作业

经济数学基础形成性考核册及参考答案

作业(一)

(一)填空题

x?sinx?___________________.答案:0

x?0xx?sinxsinxsinx?lim(1?)?lim1?lim?1?1?0 分析:limx?0x?0x?0x?0xxx1.lim?x2?1,x?02.设f(x)??,在x?0处连续,则k?________.答案:1

?k,x?0?分析:

x?0222limf(x)?lim(x?1)?limx?lim1?(limx)?lim1?0?1?1??????x?0x?0x?0x?0x?0222x?0?limf(x)?lim(x?1)?limx?lim1?(limx)?lim1?0?1?1?????x?0x?0x?0x?0x?0x?0

f(x)=limf(x)=1,所以limf(x)?1,又因为f(x)在x?0处连续, 因为lim??x?0x?0所以limf(x)?limk?k?1。

x?0x?03.曲线y?分析:y?=

x在(1,1)的切线方程是 .答案:y?12x y?|x=1 =

11x? 221 根据导数的几何意义可知曲线y?x在(1,1)的切2线斜率:k=y?|x=1 =

1 2111(x – 1),化简得:y?x? 222 由点斜式可求切线方程:y - 1=

2__.答案:2x 4.设函数f(x?1)?x?2x?5,则f?(x)?__________222分析:f(x?1)?x?2x?5?(x?2x?1)?4?(x?1)?4 f(x)?x?4 ∴f?(x)?(x?4)??(x)??4??2x 5.设f(x)?xsinx,则f??()?__________.答案:?222π2π 2分析:f?(x)?(xsinx)??x??sinx?x?(sinx)??sinx?xcosx

??(xcoxsf??(x)?(sixn?xcoxs)??(six)n)??coxs?x?coxs?x(cox)?s?coxs?coxs?xsixn?2cox

f??()?2cos()?sin()?2?0??1?

222222??????(二)单项选择题

1. 当x→+∞时,下列变量为无穷小量的是( )答案:D

1

x2?2A.ln(1+x) B. C.ex

x?1D.

sinx x分析:A.当x→+∞时,ln(1+x)→+∞

x2 B.当x→+∞时,→+∞

x?1C.当x→+∞时,e D.当x→+∞时,

?1x

2

12=()x→1

e1sinx1=?sinx→0(为无穷小) xx2. 下列极限计算正确的是( )答案:B A.limx?0xx?1 B.lim?x?0?xxxxxx?1 C.limxsinx?01sinx?1 D.lim?1

x??xx分析:A.当x→0时,lim?x?0?lim?x?0x?lim1=1 xx?0??x?lim(?1)??1 ?x?0xxx不存在。

当x→0时lim?x?0??lim?x?0当x→0时的左右极限存在,但不相等,所以limx?0B.由上面分析可知lim?x?0xx?1对。

C.limxsinx?01?0 不是等于1,而是等于无穷小(因为无空小量与有界函数乘积为x无穷小)

D.limsinx1?lim(?sinx)?0 不是等1,而是等于无穷小(因为无空小量与有

x??x??xx界函数乘积为无穷小)

3. 设y?lg2x,则dy?( ).答案:B A.

11ln101dx B.dx C.dx D.dx 2xxln10xx分析:∵y?lg2x ∴

  2xy??(lg2x)??(log10)??1111?(2x)???2x???2?

2xln102xln102xln10xln10 y??dy1   ?dy?y?dx?dx 故选:B dxxln104. 若函数f (x)在点x0处可导,则( )是错误的.答案:B

A.函数f (x)在点x0处有定义 B.limf(x)?A,但A?f(x0)

x?x0 C.函数f (x)在点x0处连续 D.函数f (x)在点x0处可微

分析:在课本第104到106页中可找到答案,具体看第104页中的 三、关于函数的连续性 第105页中的 五、关于导数、微分和连续的关系就可知道肯定B是错误的。

'5.当若f(1( ). 答案:B x)?x则f(x)?A.

1x2 B.?1x2 C.1 D.?1 xx'?21'分析: f(x)?1??x12。故选:B x....f(x)?(x)??x(三)解答题 1.计算极限

x2?3x?2(1)lim

x?1x2?1解:

(x?2)limx?lim21?2x2?3x?2(x?1)(x?2)(x?2)lim1x?1x?1x?1lim?lim?lim?????x?1x?1(x?1)(x?1)x?1(x?1)lim(x?1)limx?lim11?12x2?1x?1x?1x?1

x2?5x?6(2)lim2

x?2x?6x?8解:

(x?3)limx?lim32?31x2?5x?6(x?2)(x?3)(x?3)limx?2x?2lim2?lim?lim??x?2??x?2x?6x?8x?2(x?2)(x?4)x?2(x?4)lim(x?4)limx?lim42?42x?2x?2x?2 (3)limx?01?x?1 x解:

lim(?1)1?x?11?x?11?x?11?x?1?1?1x?0lim?lim(?)?lim?lim??x?0xxx?01?x?1x?0x(1?x?1)x?01?x?1lim1?x?lim1)1?x?0x?0

x2?3x?5(4)lim2

x??3x?2x?4解:

3?2x2?3x?5(2x2?3x?5)/x2xlim2?lim?limx??3x?2x?4x??(3x2?2x?4)/x2x??23??x2? (5)lim解:

535lim2?lim?limx??xx??x22?0?02x2?x????4243?0?03lim3?lim?limx??x??xx??x2x2sin3x

x?0sin5xlimsin3xsin3x5x13sin3x5x3sin3x1?lim(?3x??)?lim(??)??lim?limx?0sin5xx?0x?055x?0sin5x3xsin5x5x3xsin5x53x?03x5xlim13sin3x3135x?0??lim???1??sin5x553x?03x15lim5x?05x

x2?4(6)lim?4

x?2sin(x?2)解:

x2?4(x?2)(x?2)(x?2)1lim?lim?lim(x?2)??lim(x?2)??(limx?lim2)?x?2sin(x?2)x?2x?2x?2x?2sin(x?2)sin(x?2)sin(x?2)x?2lim(x?2)(x?2)1?(2?2)??41

1?xsin?b,x?0?x?2.设函数f(x)??a,x?0,

?sinxx?0?x?问:(1)当a,b为何值时,f(x)在x?0处有极限存在? (2)当a,b为何值时,f(x)在x?0处连续.

f(x)?lim(xsin解:(1)lim??x?0x?011?b)?limxsin?limb?0?b?b ??x?0x?0xxf(x)?lim lim??x?0x?0sinx?1 xx?0x?0f(x)=limf(x)=b=1,因为这里 要想使f(x)在x?0处有极限存在须有lim??没说f(x)在x?0处连续.,所以a可以取任意值。

(2)当a?b?1时,f(x)在x?0处连续。

3.计算下列函数的导数或微分: (1)y?x2?2x?log2x?22,求y? 解:

y??(x2?2x?log2x?22)??(x2)??(2x)??(log2x)??(22)??2x?2xln2?(2)y?解:

1 xln2ax?b,求y?

cx?dy??(?

ax?b(ax?b)?(cx?d)?(ax?b)(cx?d)?[(ax)??b?](cx?d)?(ax?b)[(cx)??d?])???cx?d(cx?d)2(cx?d)2a(cx?d)?(ax?b)cacx?ad?acx?bcad?bc??(cx?d)2(cx?d)2(cx?d)2(3)y?解:

13x?5,求y?

??112????y?()?[(3x?5)]??(3x?5)(3x?5)??(3x?5)2[(3x)??5?]223x?5??13??(3x?5)2?3??223112331(3x?5)32??32(3x?5)3

(4)y?解:

x?xex,求y?

1211?2111xxxy??(x?xe)??(x)??(xe)??x?[x?ex?x(ex)?]??(e?xe)??(x?1)e2212xx2xx

(5)y?esinbx,求dy 解:

axy??(eaxsinbx)??(eax)?sinbx?eax(sinbx)??eax(ax)?sinbx?eaxcosbx(bx)??eaxasinbx?eaxbcosbx?eax(asinbx?bcosbx)dy?eax(asinbx?bcosbx)dx

(6)y?e?xx,求dy

1x

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