tanx?sinx1?
x?0x3213 得tanx?sinx?x
213 即tanx?sinx~x (x?0)
2ln(1?tanx?sinx)12.lim
x?0x311.由10得lim3ln(1?12x)?limx?0x313x2 ?lim3 x?0x1?2ln(1?x2?sin3x) 13.lim
x?0tanxsinx3
x2?sinx?limx?0x2
3sinx1?2 ?lim x?0x?1
14.limx?01?xsinxe?11x2
(1?x2)2?limx?0x212x2 ?2 x1?2t nn??at?limann n??a
?t 15.limasinnnn 16.因为limx???，limy?y
n??n??ynn所以limxsinn?limy?y
n??n??xn 1..见p103例1 2.见p104例2 3.见p104例3 4.见p104例4 5.解.因为lim 习题3.11
x??1?0 x2x3?11?ax2?bx?c)2?0 所以lim(x??x?1xx3?1bc lim(2?a??2)?0
x??x(x?1)xxx3?1a?lim2?1x??x(x?1)?x3?1x3?121x??lim(?x?bx?c)?lim??x?b??0??x??x?1x??xc??x(x?1)
x3?1 b?lim?x??1x??x(x?1)x3?12lim?x?x?c?0x??x?1故c?1
6.解.因为lim1?0
x???xx??? 所以lim?x2?x?1?ax?b?1?0 x
?x2?x?1b??lim?a???0x????xx??? a?limx???x2?x?1?1xb?lim(x2?x?1?x)?0x??? 7.解.同6可得a?1
8.解.limx2sin1x2x?12x????limx???x22 ?2x221?x?31?x 9.解lim?limx??x??x?1?x?1???1?x?1??312xx?113x? x6cos25x?cos23x1?sin25x?1?sin23x-25x2?9x2 10.解lim?lim?lim?-16 222x?0x?0x?0xxx3ln1?3x?1x?11 11.解lim?lim?
x?1arcsin23x?1x?123x?12???? 12.解lim?1?3x?x?0x?02sinx?lim?1?3x?x?0x1*63x?e6
21 13.解lim2sinx?e??2x?lim?2x?x?1?x?lim?3x?1?3x?e6
*6x?0x?02sinx?x2cos12x?x2cos12?xcos12xxx 14.解lim?lim?lim??1
x?0?x?0?1?cosx?ln?x?1?x?0?1?cosx?x1?cosx?2f(x)??f(x)ln?1??f(x)f(x)sinx??limxx?2?A ?0，lim?x 15.解.因为limx?0x?0x?0a?1xa?1xlna 所以limx?0f(x)?Alna 2xxx?a?3axx-a3a?3a?x?2a?? 16.解.因为lim???lim?1??x???x?a?x???x?a? 所以a??e3a?8
ln8 32ax?12x 17.解.因为lim?lim?1
x2x?0x?02bx?x2bln(1?2x)?e?1atanx?(1?cosx) 所以a?2b
?2?ex? 18.解.左极限，因为lim4??xx?02?e? 所以
11???1 ???2?exsinx? ??lim??1?1?04x?0??2?ex??x????1?2?ex? 右极限，因为lim4x?0??2?ex????0 ???2?exsinx????0?1?1 所以lim4?x?0??2?exx???1
§4.3 连续函数的运算性质
1.?x?0,x?1且x?2?2.?x|x?0? 定义域上连续 3.?x?0? 定义域上连续 4.?x?0? 定义域上连续
定义域上连续
5.解：?f?x?在整个数轴上连续， ?e?a?0 即a?1 6.解: ?f?x?在整个数轴上连续, ?lim?x?00sinax?2 即a?2 xbx lim?1?x??f?0??2 即b?ln2 ?x?07.解：?x?1时，y?x. ?x?y?y?1?.
2 ?1?x?4时，y?x ?x?xy?1?y?16?
?x?4时，y?2 ?x?log2y?y?16? 综上所述，y?f?x?的反函数为：
?x,?x?1?,?y??x,?1?x?16?,?logx,?x?16?. ?2
§4.4 在闭区间上函数的连续性
1.证明：令f?x??x?x?1，则f?0???1?0,f?1??1?0
3 ?f?0?f?1??0 由零点定理得存在一点???0,1?使得f????0 ?方程x?x?1?0在区间?0,1?上至少有一个根。
3