数学分析习题解答1 南京廖华

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8. 由题设, ?M > 0 ? ? > 0 ?x?U ? (r, ? )? U ? (r) : | f (x) | > M / K , 故| f (x) g (x)| > M , 即x→r时f g为无穷大量.

f(x)?g(x)f(x)9. ∵lim= 1, ∴lim= 0, 即f (x) ? g (x) = o ( f (x)). x?x0g(x)x?x0f(x)

总练习题

1. (1) 2 ? x < 3时[ x] = 2, 极限为1. m(1?x???xn?1?n)?n(1?x???xm?1?m)mn??(7) , 以下参见§1?xm1?xn(1?x)(1?x???xm?1)(1?x???xn?1)

2.8(3)题的解法.

(1?a)x2?(a?b)x?(1?b)?1?a?0,2. (1) 法一. 题式=lim?a =1, b = ?1. ?0??x???a?b?0x?1?

(1?a)x2?ax?1x2?1法二. b =lim(? ax) =lim? 1 ? a = 0 ? a = 1 ? b x??x??x?1x?1

?x?1=lim= ?1. x??x?1

x2?x?1?a2x2? (2) 由题式知a < 0. 法一: b =lim(x?x?1?ax)?limx???x???x2?x?1?ax

?x?1?1. a 2 = 1? a = 1? b =limx???x2?x?1?12

法二: 0 =lim

(3) 仿(2). (1?a2)x2?(1?2ab)x?1?b2x2?x?1?ax?bx????1?a2?0, ??1?2ab?0.?

?x?2,x?2,3. 例如: (1) f (x) =? (2) f (x) = (x ? 2) ?1. x?2.?1,

?|x?x0|,x?x0,4. 例如: f (x) =?这说明局部保号性的逆命题不成立, 与局部保号性不矛 x?x0.?1,

盾.

5. 关于复合函数的极限, 有两个命题.

命题1 若limf (x ) = a,limg (y) = g (a), x0?R*, 则limg (f (x)) = g (a). x?x0y?ax?x0

证 ?? > 0 ? ? 1> 0 当0 < | y ? a | <?1 时| g (y) ? g (a)| <? . 当y = a时总有| g (y) ? g (a)| = 0 <? , 故| y ? a | <?1 时| g (y) ?g (a)| <? . 对此?1, ? ? > 0 当0 < | x ? x0 | <? 时| f (x) ? a | <?1 , 从而| g (f (x)) ?g (a)| < ? .

命题2 若limf (x) =a ,limg (y) = b, 且在x0的某去心邻域内f (x)≠a , a, b, x0?R*, x?x0y?a

则limg (f (x)) = b. x?x0

证 ?? > 0 ? ? 1 > 0 当0 < | y ? a | <? 1 时| g (y) ? b | <? . 对这个? 1, ? ? 2 > 0, 当0 < | x ? x0 | < ? 2时| f (x) ? a | <? 1 . 由条件, ? ? 3 > 0, 当0 < | x ? x0| <? 3 时| f (x) ? a | > 0 . 因此, 若取? = min {? 2, ? 3}, 则当0 < | x ? x0 | <? 时0 <| f (x) ? a | <? 1 , 故| g (f (x)) ? b | <? .

6. 2n? + ? ? ; 2n? ; (2n + 1) ? .

7. (1) 取q使r > q > 1, 则?N ?n > N : an|> q , | a n | > q n →+∞ (n→∞).

(2) 取q使r > q > 1, 则?N ?n > N : | a n | ? q | a n?1 | ? q 2 | a n?2 | ? … ? q n?N | a N |, 故 | a n | → +∞. 8. (1) an|= ( 1 + n ?1 ) n →e, 极限为+∞. (2) an|= ( 1 ? n ?1 ) n →e ?1, 极限为0.

9. (1) ? M > 0 ? N1 ?n > N1 : a n > 2M . ?N 2 ?n > N 2 : a1 + … + a n > 0. 设N ＝ N 1 + N

a1???ana1???aNaN?1???ann?Nn?N>2M > M . ??nnn?Nnn

(2) 用算术－几何平均不等式.

10. (1) 在上题(2)中取a n = n , 极限为+∞. (2) 在上题(1)中取a n = ln n, 极限为+∞.

11. 第一个等式见§3第5题. 下证f (x0 ?) = A.

先证?x?U?? (x0), f (x) ? A : 设?x'?U?? (x0)使f (x' ) > A. 由xn →x0知?xm > x', 从而 f (xm) ? f (x' ) > A. 因此n > m 时 f (xn) ? f (xm) > A, 与f (xn)→A矛盾.

? ? > 0 ?p 使 0 < A ? f (x p ) <? . 设y n ?U?? (x0), y n →x0, 则存在N, 使 y N > x p , 由f增, 得0 < A ? f (y N ) <? . 仍由f增, 得n > N时 0 < A ? f (y n ) < ? , 即lim y n = A. 由Heine定理得证.

12. 若?a > 0使f (a)≠A, 则f (a) = f (2a) = … = f (2 n a) = …, 故2 n a →∞但f (2 n a) →f (a), 与条件矛盾.

13. 若? a > 0使f (a)≠ f (1), 则 a≠1. 由 f (a) = f (a 2) = … = f (a 2 n) = …, 得 f (a 2 n)→ f (a)≠ f (1), 而1 > a > 0时a 2 n→0, a > 1时a 2 n→+∞, 与条件矛盾.

14. 设A = 0. ?? > 0 ?x 0 > 0 ?x > x0 : | f (x) ? f (x? 1) | <? . ?n ?N使x0 ? x ? n < x0 + 1. 由| f (x) ? f (x ? n) | < n? , 设M是f在[x0 , x0 +1)上的界, 则有| f (x)| ? | f (x ? n)| + n? < M + x?, | f (x) / x | < M / x + ? < 2? , 只要x > M / ? .

设A≠0. 令 g (x) = f (x) ? Ax, 则 g (x + 1) ? g (x)→0, 故g (x) / x→0, f (x) / x→A . 2 , 则n > 2N时

第四章函数的连续性

§1 连续性概念

6. 由第三章§3第5题或定理3.10, 在间断点处, 左右极限总是存在的.

7. g (a) =limf(y), 故?? > 0 ?? > 0, 当0 < | y ? a | 时 | f (y) ? g (a) | < ? ? . 因此 | y?a

x ? a | <? 时| g (x) ? g (a) | = lim| f (y) ? g (a) | ? ? ? < ? , 即limg (x) = g (a), g连续. y?xx?a

111111,,时f (x) = 1, x =,,时f (x) = 0. 234234

(2) 例如: f (x) = D (x) (x ? ? )(x ? 1/ 3) (x ? 1/ 4), 其中D为Dirichlet函数.

11(3) 例如: x≠时f (x) = 1, x =时f (x) = 0. nn

(4) 例如: x为0, 正有理数及无理数时f (x) = x D (x), x为负有理数时f (x) = 1.

§2 连续函数的性质

17. 设F (x) = f (x + a) ? f (x) (x?[0, a]), 则F在[0, a]上连续, F (a) F (0) = ? ( f (a) ? f (0) )2 ? 0. 于是, 当 f (a) = f (0) 时取x 0 = 0, 当 f (x)≠ f (0) 时应用介值定理得证.

18. 由第三章§3第5题, ?c ?[a, b), f (c) ? f (c+). 若 f (c) < f (c+), 则 ( f (c), f (c+)]将不含于f的值域[ f (a), f (b)], 因此 f (c) = f (c+), f 在c右连续. 类似地, 由 f (d?) ? f (d)可证f在d?(a, b]左连续.

19. 设n个数 f (x1), …, f (xn)中最小的是f (xi), 最大的是f (xk), 则f (xi) f(x1)???f(xn)f(x1)???f(xn)?? f (xk), 由介值定理, 有xi , xk间的? 使 f (? ) = . nn

20. (上接提示) ?? > 0 ?? =? , 当| x' ? x" | <? 时|cosx?? cosx??| <? , 即f 在[1,∞)上一致连续. 又, f在[0,1]上一致连续, 故f在[0,∞)上一致连续. 9. ( 1) 例如: x≠

总练习题

1. 设函数F如下: F (x) = f (x) (x?(a, b)), F (a) = f (a +), F (b) = f (b ?), 则F在[a, b]上连续.

(1) F在[a, b]上有界, 故f在(a, b)内有界.

(2) F有最大值, 设为 F (c). 若 F (c) = f (? ), 则f (? ) 即为所求. 若 F (c) > f (? ) ? max { f (a +), f (b ?)} = max { F (a), F (b)}, 则c≠a, b, c?(a, b), f (c)即为所求.

2. 任取x 0 ?(a, b). 由f (a+) = f (b?) = +∞, ? c, d?(a, b) 使 c < x0 < d 且 x?(a, c)∪(c, b) 时 f (x) ? f (x 0), 则f 在[c, d]上的最小值m就是f在(a, b)上的最小值: ?x?(a, b), x?[c, d] 时f (x) ? m, x?(a, c)∪(c, b) 时 f (x) ? f (x 0) ? m.

3. (1) 对x?I, 取有理数列rn →x, 则 f (x) = lim f (r n) = 0.

(2) 设x1, x2 ?I, x1 < x2 . 依次在[x1, ? (x1 + x2)), ( ? (x1 + x2), x2]中取有理数列{rn}, {sn}使rn↓x1 , sn↑x2 , 则x1 < rn <…< r1 < s1 <…< s n < x2 , f (x1) = lim f (r n) ? r 1 < s1 ? lim f (sn ) = f (x2), f 严格增.

4. 设方程左端为 f (x), 则 f (? 1 +) = +∞, f (? 2 ?) = ?∞, 故? a?(?1, ? (?1 + ?2)), b?( ? (?1 + ?2), ?2) 使 f (a) > 0, f (b) < 0, 从而?? ?(a, b)? (? 1, ?2)使 f (? ) = 0.

5. 法一. 用书上提示. | f |在[a, b]上连续, 故有最小值m = | f (? )|. 若m = 0, 则已得证.若m > 0, 则有y?[a, b]使| f (y)| ? ? | f (? )| < m, 与m是最小值矛盾.

法二. 任取x1?[a, b]. 由题设, ?x2?[a, b]使| f (x2)| ? ? | f (x1)|,?x3?[a, b]使| f (x3)| ? ? | f (x2)|, 如此继续, 得[a, b]中的数列{xn}, 满足| f (xn)| ? ? | f (xn?1)| ? … ? 2 1?n | f (x1)|. {xn}有界, 有收敛子列, 设为{xnk}, 且收敛于? , 则??[a, b], | f (xnk)| ?21?nk| f (x1)|. 令k→∞, 得| f (? )| ? 0, 即 f (? ) = 0.

6. 仿§2第19题.

7. (1) 0 ? f (an) ? an = f (an?1) ? an?1, {an}减且有下界0, 故收敛.

(2) 在an = f (an?1)中令n→∞.

(3) 由(2), t ? 0. 若t > 0, 则0 ? f (t) < t. 但类似于(2)可得f (t) = t, 矛盾.

8. n = 1时? = 0满足要求. 设n > 1, F (x) = f (x + n ?1) ? f (x), 则F在[0, 1 ? n ?1]上连续. 若??[0, 1 ? n ?1], F (x)≠0, 则F > 0或F < 0. 设F > 0, 即??[0, 1 ? n ?1]有 f (x) < f (x + n?1), 则f (0) < f (n?1) < f (n?2) <…< f (1), 与条件矛盾.

9. (1) 由f (0 + 0) = 2 f (0)得f (0) = 0.?a?R, 当x→a即x ? a →0时, f (x) ? f (a) = f (x ? a)→f (0) = 0, 即f在a连续.

(2) ?m, n?N由条件有 f (n x) = n f (x), f (x) = m f (x / m), f (x / m) = f (x) / m , f (n x / m) = (n / m) f (x), 即对正有理数r有f (r x) = r f (x). 又, 由0 = f (0) = f (x) + f (?x)得 f (x) = ? f (?x), 故对所有有理数r有f (r x) = r f (x), 特别地, 有 f (r) = r f (1). ?x?R, 取有理数列rn →x (n→∞), 则 f (xn) = rn f (1), 令n→∞得证.

10. x > 0时 f (x) =f(x)???f(x2?n). 令n→∞得 f (x) = f (1). x < 0时x 2 > 0, 故 f (x) = f (x 2 ) = f (1). 又, f (0) =limf (x) =limf (1) = f (1). x?0x?0

第五章导数和微分

§1 导数的概念

10. 设f?' (x0) = c, 则x →x 0 ? 时 f (x ) = f (x 0) + c (x ? x 0) + o (x ? x 0)→f (x 0), 即f在x0左连续. 同理, f 在x0 的右导数存在时右连续, 故左右导数都存在时连续.

便得 f ' (x) = f (x) f ' (0) = f (x). 下证 f (0) = 1: x1 = x2 = 0? f (0) = f 2 (0) ? f (0) = 0, 1; 而f (0) = 0 ? f (x) = f (x + 0 ) = f (x) f (0) = 0 ? f ' (x) = 0 ? f ' (0) = 0, 与条件矛盾. g(x)1g(x)g(x)?g(0)sin|?||?||→g' (0) = 0, xxxxf(x)g(x)1∴f ' (0) = lim?limsin= 0. x?0x?0xxxf(x?h)?f(x)f(x)f(h)?f(x)f(h)?112. f ' (x) =lim=lim= f (x)lim. 若f (0) = 1, h?0h?0h?0hhh11. ∵|

f(x0??x)?f(x0)f(x0)?f(x0??x)1?)?( f ' (x0) + f ' (x0)) = 右. ?x?02?x2?x2

?(b) > 0???2 ? x?(b ??2, b) : f (x) 14. f+' (a) > 0? ??1 ? x?(a, a +?1) : f (x) > f (a) = K.f?13. 左 =lim(

< f (b) = k. 因此 ?c, d?(a, b)使f (c) < K < f (d ), 应用介值定理得证.

15. 取铁链最低点为原点, 可设铁链方程为y = f (x) = k x 2 . 由x = 50, y = 10得k = 1 / 250. 设所求角为?, 由于支点之一的横坐标为50, 故tan (?? ? ? ) = f ' (50) = 2 / 5, ? = ? ? ? arctan (2 / 5).

16. 设P (x 0 , y 0 ), Q (x 1 , y 1 ,), 则过的切线方程为y ? y 0 = 3 x 0 2 (x ? x 0 ), 与y = x3联立, 解得x 1 = ? 2 x 0 , 故3x 12 = 4 ? 3x 02 , 即y ' |Q = 4y ' |P .

§2 求导法则

6. 条件? 2x f ' (x 2) = 2 f (x) f ' (x)? f ' (1) = f (1) f ' (1)? f ' (1) = 0或 f (1) = 1.

§3 参变量函数的导数

4. y ' = tan t, (x, y)处的法线方程为Y = y ? cot t (X ? x), 故原点到法线的距离

|xcott?y|d == | y sin t + x cos t | = | a |. ?cot2t

5. x = r cos ? = a sin 2? , y = r sin? = 2a sin 2? . y ' = tan 2? , 即任一点的切线与x轴正向的夹角为2? , 而向径与x轴正向的夹角为? , 故切线与向径的夹角为? .

6. x = r cos? = a ( 1 + cos? ) cos? , y = a ( 1 + cos? ) sin ? ,

dya(cos2??cos?)3?3?????cot?tan(?). dx?a(sin2??sin?)222

设所求夹角为? , 则? = 3?????????. 2222

§4 高阶导数

2. (f (x 2))' = ( f 2 (x))"? 2x f ' (x 2) = (2 f (x) f ' (x))' = 2 ( f ' 2 (x) + f (x) f " (x)). 当x = 1时左端=2 f ' (1) = 0, 右端 = 2 f ' 2 (1) +f (1) f " (1) = 0.

7. f (x) = (x sgn x) 3 = x 3 sgn x . x≠0时 f ' (x) = 3x 2 sgn x, f " (x) = 6x sgn x , f '" (x) = 6 sgn x, 更高阶的导数为0. x = 0时由定义或导数极限定理(第六章§1)可得f ' (0) = f " (0) = 0,更高阶的导数不存在. 1d()f?(x)dxf??(x)dx11d2xddxd11??.2?????8. dydyf?(x)dydydydyf?(x)dxdy(f?(x))2f?(x)

f??(x)f??(x)dxf?(x)f???(x)?3(f??(x))2d3xdd2xd?. ??(?)??. (f?(x))3dy3dydy2dx(f?(x))3dy(f?(x))5

9. (1) (1 + x 2 ) y' = 1, 求导得( 1 + x 2 ) y" + 2x y' = 0.

(2) (1)中的方程两端求n阶导数, 得(1 + x2) y (n+2) +2x (n + 1) y (n+1) + n (n +1) y (n) = 0, ∴y (n+2) (0) = ?n (n ? 1) y (n) (0). ∵y' (0) = 1, y"(0) = 0, ∴y (2k) (0) = 0, y (2k+1)(0) = (?1)k (2k)!. 10. (1) ?x2y' = 1, 求导, 得( 1 ? x 2 ) y" = x y' . 再求n阶导数, 得( 1 ? x 2 ) y (n+2) ? 2nx y (n+1) ? n ( n ? 1) y (n) = x y (n+1) + n y (n), 即( 1 ? x 2 ) y (n+2) ? (2n + 1) x y (n+1) ? n 2 y (n) = 0.

(2) (1)的结果中令x = 0, 得y (n+2) (0) = n 2 y (n) (0), 故y (2k) (0) = 0, y (2k+1)(0) = ( (2k?1)!!)2.

?1?2exp(?x?2)?0x?x?lim?lim11. 法一 f ' (0) =lim= 0 . x?0x?0exp(x?2)x?0?2x?3exp(x?2)x

?2x≠0时f ' (x) = 2e?x?2f (x), 故?x?R, x 3 f ' (x) = 2 f (x). 求n阶导数, 得 x3x3

2 f (n) (x)= x 3 f (n+1) (x) + n23x2 f (n) (x) +n(n?1)n(n?1)(n?2)· 6x f (n?1) (x) +·6 f (n?2) (x). 62

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