高等数学基础形考作业1答案:
第1章 函数 第2章 极限与连续
(一)单项选择题
⒈下列各函数对中,(C)中的两个函数相等.
2 A. f(x)?(x),g(x)?x B. f(x)?x2,g(x)?x
x2?1 C. f(x)?lnx,g(x)?3lnx D. f(x)?x?1,g(x)?
x?13⒉设函数f(x)的定义域为(??,??),则函数f(x)?f(?x)的图形关于(C)对称. A. 坐标原点 B. x轴 C. y轴 D. y?x ⒊下列函数中为奇函数是(B).
A. y?ln(1?x) B. y?xcosx
2ax?a?x C. y? D. y?ln(1?x)
2 ⒋下列函数中为基本初等函数是(C). A. y?x?1 B. y??x C. y?x2??1,x?0 D. y??
1,x?0?⒌下列极限存计算不正确的是(D).
x2?1 B. limln(1?x)?0 A. lim2x??x?2x?0sinx1?0 D. limxsin?0
x??x??xx⒍当x?0时,变量(C)是无穷小量.
sinx1 A. B.
xx1 C. xsin D. ln(x?2)
x C. lim⒎若函数f(x)在点x0满足(A),则f(x)在点x0连续。
A. limf(x)?f(x0) B. f(x)在点x0的某个邻域内有定义
x?x0 C. lim?f(x)?f(x0) D. lim?f(x)?lim?f(x)
x?x0x?x0x?x0
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(二)填空题
⒈函数f(x)?x2?9?ln(1?x)的定义域是?3,???.
x?32⒉已知函数f(x?1)?x?x,则f(x)? x2-x .
1x⒊lim(1?)?e2. x??2x1?x?⒋若函数f(x)??(1?x),x?0,在x?0处连续,则k? e .
?x?0?x?k,1⒌函数y???x?1,x?0的间断点是x?0.
?sinx,x?0⒍若limf(x)?A,则当x?x0时,f(x)?A称为x?x?x0x0时的无穷小量。
(三)计算题
⒈设函数
?ex,x?0 f(x)???x,x?0求:f(?2),f(0),f(1).
解:f??2???2,f?0??0,f?1??e?e
1⒉求函数y?lg2x?1的定义域. x?2x?1??x?0??2x?11?解:y?lg有意义,要求?解得?x?或x?0
x2??x?0???x?0? 则定义域为?x|x?0或x???1?? 2?⒊在半径为R的半圆内内接一梯形,梯形的一个底边与半圆的直径重合,另一底边的两个端点在半圆上,试将梯形的面积表示成其高的函数. 解: D A R O h E
B C
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设梯形ABCD即为题中要求的梯形,设高为h,即OE=h,下底CD=2R 直角三角形AOE中,利用勾股定理得
AE?OA2?OE2?R2?h2 则上底=2AE?2R2?h2
hg2R?2R2?h2?hR?R2?h2 2sin3x⒋求lim.
x?0sin2xsin3xsin3x?3xsin3x3133解:lim?lim3x?lim3x?=??
x?0sin2xx?0sin2xx?0sin2x2122?2x2x2x故S?????x2?1⒌求lim.
x??1sin(x?1)x2?1(x?1)(x?1)x?1?1?1?lim?lim???2 解:limx??1sin(x?1)x??1sin(x?1)x??1sin(x?1)1x?1tan3x.
x?0xtan3xsin3x1sin3x11解:lim?lim??lim??3?1??3?3
x?0x?0xxcos3xx?03xcos3x1⒍求lim1?x2?1⒎求lim.
x?0sinx1?x2?1(1?x2?1)(1?x2?1)x2?lim?lim解:lim2x?0x?0x?0sinx(1?x?1)sinx(1?x2?1)sinx ?limx?0
x(1?x2?1)sinxx?0?0
?1?1??1⒏求lim(x??x?1x). x?31?111(1?)x[(1?)?x]?1x?1xe?1xxx?x解:lim()?lim()?lim?lim?3?e?4 xx??x?3x??x??x??33e11?(1?)x[(1?)3]3xxx3x2?6x?8⒐求lim2.
x?4x?5x?4x?4??x?2?x2?6x?8x?24?22?解:lim2?lim?lim??
x?4x?5x?4x?4?x?4??x?1?x?4x?14?13
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