高等数学第一章总习题及答案

7. 已知 lim
x →0
f ( x) ) sin x = 3 , 求 lim f ( x) . x →0 x 2 2x − 1
解
因为 lim(2 − 1) = 0 , lim
x →0
x
ln(1 +
x →0
f ( x) ) sin x = 3 , 故必有 lim ln(1 + f ( x) ) = 0 , x →0 sin x 2x − 1
2
2
x
1 1 . = ( )2 = 2 2
1
(4) (5) (6)
lim
x →0
1 x sin x 1 = lim 2 2 = . x →0 2 x
1
x
lim(1 + 3tan 2 x)cot
x →0
= [lim(1 + 3tan 2 x) 3tan x ]3 = e3 .
2
x →0
设 k 为任一个大于 2c 的自然数, 则当 n > k 时,
0 < x ≤ e, 在 x = e 处, lim+ f ( x) = ln e = 1 , lim− f ( x) = 1 , x →e x →e x > e,
故 f ( x) 在 x = e 处连续, 故函数连续区间为 (0, + ∞) .
9.
⎧ cos x , x ≥ 0, ⎪ ⎪x + 2 设 f ( x) = ⎨ 要使 f ( x) 在 (−∞, + ∞) 内连续, 应如何选择 ⎪ a − a − x , x < 0, ⎪ x ⎩
n →∞ n →∞
(B) 无界数列必定发散; (D) 单调数列必有极限.
yn . xn
3. 设 lim xn = +∞ , lim yn = y ( y ≠ 0) , 求 lim xn sin
n →∞
解
y lim xn sin n = lim n →∞ xn n →∞
sin
yn xn
yn xn
⋅ yn = y .
2 cos x+a x−a x−a sin ⋅ sin 2 2 = lim cos x + a ⋅ lim 2 x→a 2 x→a x − a x−a 2
n →∞
解
(1)
sin x − sin a lim = lim x→a x→a x−a
= cos (2)
a+a ⋅ 1 = cos a . 2
3x − 4 x + x − 1 + x2 − 2x + 3
2
x →∞
= 0 , 故必有 a 2 = 1 , 且
x →+∞
lim ( x 2 − x + 1 − ax − b) = lim 1 . 2
ln(1 +
−(1 + 2ab) x + (1 − b 2 ) x − x + 1 + ax + b
2
x →+∞
=
−(1 + 2ab) =0, 1+ a
故 a = 1, b = −
解
6. (1) (2)
确定常数 a 及 b 的值, 使下列极限等式成立.
x →∞
lim (
x + 2a x ) =8; x−a
x →∞
lim ( x 2 − x + 1 − ax − b) = 0 .
lim x + 2a x 3a x 3a 3a ⋅ x − a ) = lim (1 + ) = lim (1 + ) = e x→∞ x − a = e3a = 8 , x →∞ x →∞ x−a x−a x−a x − a 3ax 3ax
因为 2 x − 1 ∼ ln 2 ⋅ x , ln(1 +
f ( x) f ( x) , 所以 )∼ sin x sin x
3
ln(1 + lim
x →0
f ( x) f ( x) ) sin x = lim sin x = 1 lim f ( x) = 1 lim f ( x) = 3 , x x →0 ln 2 ⋅ x ln 2 x →0 x sin x ln 2 x →0 x 2 2 −1
0<
k
cn c c =( ⋅ n! 1 2
c c c )( ⋅ k k +1 k + 2
c 1 (2c) ) < c k ⋅ ( )n−k = n , n 2 2
k
由于 lim
(2c) cn = 0 , 由夹逼准则, 故 lim =0. n n →∞ n ! n →∞ 2 n 1 (1 − a )(1 + a )(1 + a 2 ) (7) lim (1 + a )(1 + a 2 ) (1 + a 2 ) = lim n →∞ n →∞ 1 − a =
(2)
x ln[en (1 + ( ) n ] ln(e + x ) e 当 0 < x < e 时, f ( x) = lim = lim n →∞ n →∞ n n
n n
x x ln(1 + ( ) n ) ( )n e ] = lim[1 + e ] = 1 ; = lim[1 + n →∞ n →∞ n n
解
(1) lim (
x →∞
所以 a = ln 2 .
(2)
x →∞
lim ( x 2 − x + 1 − ax − b) = lim
( x 2 − x + 1) − (ax + b) 2 x 2 − x + 1 + ax + b
x →∞
= lim
(1 − a 2 ) x 2 − (1 + 2ab) x + (1 − b 2 ) x − x + 1 + ax + b
=
1 2 a
故 a = 1. 10. 设常数 a > 0 , b > 0 , 证明方程 x = a sin x + b 至少有一个正根, 并且它不超 过a +b. 证 令函数 f ( x) = ( x − b) − a sin x , f (0) = (0 − b) − a sin 0 = −b < 0 , f (a + b) = (a + b − b) − a sin(a + b) = a − a sin(a + b) , 当 sin(a + b) < 1 时 , f (a + b) > 0 , 由零点定理 , 至少存在一点 ξ ∈ (0, a + b) , 使 得 f (ξ ) = 0 , 即 ξ 为原方程的根, 它是正根且不超过 a + b ; 当 sin(a + b) = 1 时, f (a + b) = 0 , 则 a + b 是原方程的正根, 且不超过 a + b . 11. 设函数 f ( x) 在 [0, 2a ] 上连续 , 且 f (0) = f (2a ) , 证明 : 在 [0, 2a ] 上至少存 在一点 ξ , 使 f (ξ ) = f (ξ + a ) . 证 构造函数 F ( x) = f ( x) − f ( x + a ) , 则 F ( x) 在 [0, a ] 上连续, 且 F (0) = f (0) − f (a ) , F (a ) = f (a ) − f (2a ) = f (a ) − f (0) , 若 f (0) = f (a ) , 则 ξ = 0 即是满足 f (ξ ) = f (ξ + a ) 的点； 若 f (0) ≠ f (a) , 则必有 F (0) 与 F (a) 异号, 故由零点定理, 至少存在一点 ξ ∈ (0, a) , 使得 F (ξ ) = 0 , 即 f (ξ ) = f (ξ + a) . 综上, 至少存在一点 ξ ∈ [0, a ] ⊂ [0, 2a] , 使 f (ξ ) = f (ξ + a ) .
2
x →+∞
lim ( x 2 + x − 1 − x 2 − 2 x + 3) = lim
x →+∞
= lim
3− 1+
4 x
x →+∞
1 1 2 3 − 2 + 1− + 2 x x x x
1− x 1
=
3 . 2
(3)
1 − x 1− x 1 1+ lim( ) = lim( ) x →1 1 − x 2 x →1 1 + x 1 + x sin x − 1 ex − 1
2
1− x
(4)
lim
1 + x sin x − 1 ex − 1
2
x →0
;
(5) (7) (8)
(6)
n
cn (c > 0) ; n →∞ n ! lim
n →∞
lim (1 + a)(1 + a 2 ) lim ( 1 2 + + 2! 3! +
(1 + a 2 ), a < 1 ; n ). (n + 1)!
当 x = e 时, f ( x) = lim
ln(en + x n ) ln(2en ) = lim =1; n →∞ n →∞ n n e e ln[ x n (1 + ( ) n ] ln(1 + ( ) n ) x x ] = lim[ln x + n →∞ n n
ln(en + x n ) 当 x > e 时, f ( x) = lim = lim n →∞ n →∞ n e ( )n = lim[ln x + x ] = ln x . n →∞ n ⎧1, 即 f ( x) = ⎨ ⎩ln x,ຫໍສະໝຸດ 故 limx →0
f ( x) = 3ln 2 . x2
8. (1)
写出下列函数的连续区间与间断点, 并指出间断点的类型:
f ( x) = x 2 − 1 x −1 e ; x −1
1
(2)
解
ln(en + x n ) ( x > 0) . n →∞ n (1) 易知连续区间为 (−∞, 1) ∪ (1, + ∞) , 因为 f ( x) = lim
n +1 1 1 lim (1 − a 2 ) = . n →∞ 1− a 1− a
(1 + a 2 )
n
2
(8)
n →∞
lim (
1 2 + + 2! 3!
+
n 1 1 1 1 ) = lim[( − ) + ( − ) (n + 1)! n→∞ 1! 2! 2! 3!
+(
1 1 )] − n ! (n + 1)!
x x + arcsin , 则它的定义域是 3 x−2
x → x0 x → x0
[−3, 0) ∪ (2,3]
;
(3)
若 f ( x) < g ( x) , 且 lim f ( x) = A , lim g ( x) = B , 则 A 和 B 的关系是
A≤ B ; (4)
要条件是 解 设函数 f ( x) 在点 x0 的某邻域内有定义, 则 f ( x) 在 x = x0 处连续的充分必
数a?
4
解
易知函数在 (−∞, 0) ∪ (0, + ∞) 上连续, 要是函数在 x = 0 处也连续, 则
x →0
lim− f ( x) = lim−
x →0
a − a−x x 1 = lim− = lim− x →0 x( a + a − x ) x →0 x a + a−x = f (0) = 1 , 2
第一章总习题
1. (1) 填空题:
⎧e − x , x ≤ 0, ⎪ 设 f ( x) = ⎨ 则 f (−1) = ⎪ ⎩cos x, x > 0,
x −1 ⎧ ⎪e , f (1 − x ) = ⎨ 2 ⎪ ⎩cos(1 − x ),
2
e
,
2
x ≥ 1, x < 1;
(2)
设函数 f ( x) = lg
4. 求下列极限: (1) sin x − sin a ; x →a x−a lim (2)
x →+∞
lim ( x 2 + x − 1 − x 2 − 2 x + 3) ;
1
(3)
1 − x 1− x lim( ) ; x →1 1 − x 2 lim(1 + 3tan 2 x)cot x ;
x →0
1 1 ) = 1. = lim ( − n →∞ 1! ( n + 1)!
1
5.
已知当 x → 0 时, (1 + ax 2 ) 3 − 1 与 cos x − 1 为等价无穷小, 求数 a .
1 1 2 ax 3 (1 + ax 2 ) 3 − 1 2 3 由已知, lim = lim = − a =1, 故 a = − . 1 x →0 x → 0 2 cos x − 1 3 − x2 2
1 1 1 1
x 2 − 1 x −1 x 2 − 1 x −1 lim+ e = lim+ ( x + 1)e x −1 = +∞ , lim− e = lim− ( x + 1)e x −1 = 0 , x →1 x − 1 x →1 x →1 x − 1 x →1
故 x = 1 是第二类间断点.
+ − f ( x0 ) = f ( x0 ) = f ( x0 )
.
(1) 略.
由
(2) (3) (4) 2. (A) (C) 解
x x > 0 , 且 −1 ≤ ≤ 1 得, 函数定义域为 [−3, 0) ∪ (2,3] . 3 x−2
略. 略. 下列四个命题中正确的是( B ). 有界数列必定收敛; 发散数列必定无界; 略.