2.解
取x k =
1 , 则f ( x k ) = ( 2kπ ) 2 sin(2kπ ) = 0.故x → ∞时，f ( x )不是无穷大， 2 kπ
1 1 1 排除( B)；显然x → 0时， 2 sin 不是无穷小，排除( A )；取xk = ,则 π x x 2 kπ + 2 π π 1 1 f ( xk ) = ( 2kπ + ) 2 sin(2kπ + ) → ∞， 2 sin 不是有界，排除(C). x x 2 2 所以应选( D)
a 2x − 1 lim ln y = lim ln = lim = = x →1 x →1 x − 1 x →1= x 所以 lim e ln y = e x →1+ y = lim = =
x →1 x →1 x →1 lim ln y
= ea .
x →1 x →1 x →1
当a = ln 2时，e a = 2, 即 lim f ( x ) = lim f ( x ) = lim f ( x )，此时 lim f ( x )存在 − + 所以应选( D).
n →∞
令 lim = a , 对 x n +1 = 6 + x n 两边取极限得 a = 6 + a , 从而 a 2 − a − 6 = 0 . 因此，
n→∞
a = 3 或 a = −2. 因为 x n > 0 ( n = 1 , 2 , Λ ) , 所以 a ≥ 0 . 舍去 a = −2 , 故得极限值为 a = 3.
3.解 由于f ( x )在x = 0处可导，由导数定义知 lim
x →0
f ( x ) − f ( 0) = f ′(0) x
现f (0) = 0, f ′(0) ≠ 0, 所以 f ( x) f ( x) lim+ F ( x ) = lim+ = f ′(0) ≠ f (0), lim− F ( x ) = lim− = f ′(0) ≠ f (0) x →0 x →0 x →0 x →0 x x 因此x = 0是F ( x )的第一类间断点.所以只有( B)项正确.
x
π
2
= lim
x →1
cos( x − 1) − 1 1 − sin
x =− x 4 4
π
2
x
x
π
π
2
= lim x
−1
π
2
4
sin
π
2
π2
.
故f ( x )在x = 1处不连续，若使之连续，需定义f (1) = −
π2
3பைடு நூலகம்解
π 1 f ' ( x ) = a (sin x + x cos x )，故 f ' ( ) = a，所以 a = . 2 2
8.解 x = 0为第一类间断点， x = 1为第二类间断点.
二、选择题
1.解 f ( x ) = lim lim − −
x →1
x −1 1− 2 − x
x →1
= lim −
x →1
0 0
1 1 2 2−x
= 2,
令y = (
2x − 1 ) x
a x −1
, 则 ln y =
a 2x − 1 ln , x −1 x 1 1 1− a ln[1 + (1 − )] x = lim a ⋅ x = a, = x → 1 x −1 x −1
f ( x )＝ lim f ( x ) = f (1), 4.解 要使f ( x )在x = 1处可导，则f ( x )必须在x = 1处连续，即 lim ＋ −
x →1 x →1
f ' ( ) = lim Δx → 0 2
π
f(
π
+ Δx ) − f ( ) 2 2 = 1, Δx 2
π
也即：a + b = 0 ⇒ b = − a. f ( x )在x = 1处可导，则必有 f − ' (1) = f + ' (1) x sin[ ( x − 1)] π 2 2 =− f − ' (1) = lim lim = − − x →1− x → 1 x −1 x −1 2 ax 2 + b − ( a + b) a ( x 2 − 1) f + ' (1) = lim = lim = 2a x →1＋ x →1＋ x −1 x −1 x cos x − ( a + b) 所以 2a = −
x →1 x →1 x →1 a
−e −2=0
(e − 2)(e + 1) = 0 ⇒ e = 2 ⇒ a = ln 2
a a a
7.解
dy = 2 xf ' ( x 2 ) cos[ f ( x 2 )] dx
d2 y = 2{f ' ( x 2 ) cos[ f ( x 2 )] + 2 x 2 f " ( x 2 ) cos[ f ( x 2 ) − 2 x 2 [ f ' ( x 2 )] 2 sin[ f ( x 2 )} 2 dx = 2 f ' ( x 2 ) cos[ f ( x 2 )] + 4 x 2 { f " ( x 2 ) cos[ f ( x 2 )] − [ f ' ( x 2 )] 2 sin[ f ( x 2 )]}
Δx → 0
5.解 根据定义 lim
6.解 ∴e
2a x →1
o ( Δx ) Δy − dy ＝ lim = 0. Δ x → 0 Δx Δx lim f ( x ) = lim 3x = 3 lim f ( x ) = lim (e 2 ax − e ax + 1) = e 2 a − e a + 1 − − + +
1.证 由 x1 = 10 及 x 2 = 6 + x1 = 4 知 x1 > x 2 设对某正整数 k 有 x k > x k +1， 则有 x k +1 = 6 + x k > 6 + x k +1 = x k + 2 故由归纳法知，对一切正整数 n , 都有x n > x n +1 . 即 {x n }为单调减少数列. 由 x n +1 = 6 + x n , 显见， x n > 0 ( n = 1 , 2 , Λ ) , 即 {x n } 有下界。根据单调有界数列 必有极限知 lim x n 存在.
x → 0时，显然f ( x )是一个无穷小量，比较f ( x )与x的阶数，需要根据极限 f ( x) lim 的值进行判别，这里只须知e x − 1 ~ x ( x → 0)就能判别本题，因为 x →0 x 2 x + 3x − 2 e x ln 2 + e x ln 3 − 2 e x ln 2 − 1 e x ln 3 − 1 lim = lim = lim + lim x →0 x →0 x →0 x →0 x x x x x ln 2 x ln 3 e e −1 −1 = lim ⋅ ln 2 + lim ⋅ ln 3 = ln 2 + ln 3 ≠ 1 → x →0 x 0 x ln 2 x ln x 故f ( x )与x是同阶但非等价无穷小量，所以只有( B)项正确. 4.解
答案:本科高等数学作业卷测试题(一)
一、填空题
1.解
t ⎡ ⎤ x x x ⎛ ⎞ ⎛ ⎞ f ( x ) = lim⎜1 + ⎟ = ⎢lim⎜1 + ⎟ ⎥ t →∞ ⎢ t →∞ ⎝ t⎠ t⎠ ⎥ ⎝ ⎣ ⎦ 2t 2x
= e 2 x , 则 f (ln 2) = 4
2.解 根据无穷小量与有界函 数之积仍为无穷小量知 应填0.
2
x →0 −
lim f ( x ) = lim− x 2 g ( x ) = 0 = f (0)
x →0
1 − cos x f +′ (0) = lim x x
x →0 + 0
−0
1 2 x = lim 2 3 = 0
x →0+ 0
x2
∴ f ( x )在x = 0处可导，故应选( D).
三、计算题
1 2 x 1 − cos x = lim+ 2 = 0 = f (0), 5.解 lim= f ( x ) = lim+ x →0 x →0 x →0 x x ( A), ( B)不正确. 故f ( x )在x = 0处极限存在，且连续， x g ( x) − 0 f −′ (0) = lim = lim xg ( x ) = 0, x →0− 0 x →0−0 x
2.解 lim f ( x ) = lim
x →1 x →1
ln cos( x − 1) 1 − sin
x →1
π
= lim
x →1
ln[1 + cos( x − 1) − 1] 1 − sin
x →1
= lim
x →1
− sin( x − 1) −
π
2
cos
π
2
= lim −
2 − x +1 2 cos
π
π
π
2
⇒ a=−
π
4
. 从而 b =
π
4
.
5.解
dy dy = dx d t
dx tf ' ' (t ) = = t, d t f ' ' (t ) x arctan
d2 y d dx 1 = (t ) = . 2 dt d t f ′′(t ) dx