tan原式=limn???4?tan1n(1?tan2?4?tann)=
lim1n??(1?tan(1?tan1n1nn)nn)
=
ee?1=e
5．求函数f(x)?limx(1?sin?x)?sin?x1?(1?sin?x)12nnn??,?1?x?1
12 解：（1）显然有f(0)?0,f(1)? , f(?1)??
（2）当0?x?1时，1?sin?x?1,有lim(1?sin?x)n??
n??所以，f(x)?x
n （3）当?1?x?0时，0?1?sin?x?1, 有lim(1?sin?x)?0
n??所以，f(x)?sin?x
1??,x??1?2?sin?x,?1?x?0?从而，f(x)??
x,0?x?1?1?,x?1?2?6．设limf(x)存在，f(x)?3x?2xlimf(x)，求f(x)
x?12x?12解：令limf(x)?l，则f(x)?3x?2xl，limf(x)?lim(3x?2xl)?3?2l?l?l??3
2x?1x?1x?1故f(x)?3x?6x
127．求函数f(x)?2x?11的不连续点且判别类型。
12x?1解：显然x=0为间断点。因为lim?f(x)?lim?x?0x?012x?11?1,
2x?1又lim?f(x)?lim?x?0x?02x?11??1，故x=0为第一类间断点（跳跃间断点）。
2x?1
1??f(a?)??n8．已知f(x)在x=a处可导，且f(x)?0,n为自然数，求lim?? x???f(a)?????1??f(a?)??n解：lim??x??f(a)??????nn1??f(a?)?f(a)?1?n?lim?1??? x??1nf(a)????n??f(a)f(a??1n)?f(a)nnf(a)1n1??f(a?)?f(a)?1?n ?lim?1???x??1nf(a)????n??f(a?)?f(a
f(a?1 因为limn??nf(a))?f(a)n?limn??f(a?1n1)?f(a)?1f(a)?f?(a)f(a)
n1??f(a?)??n所以，lim??x??f(a)??????nf?(a)?ef(a)
9．设y?f(3x?23x?2dxdyd3x?23x?23x?23x?2212解： ?f()?f?()?()??arcsin()?2dxdx3x?23x?23x?23x?2(3x?2)2),f?(x)?arcsinx，求
dyx?0
于是，
dydxx?0?(arcsin1)?3?32?
2?x?cos(t)2dydy?2t1,10．设?，求 22y?tcos(t)??cosudu,t?0dxdx?12u?12222222cos(t)?2t??2tsin(t) 解：x???2tsin(t),y??cos(t)?2tsin(t)?2tdydx?t ，
dydx22?ddx(dydx)?ddx(t)??12tsin(t)2
23x?f?(t)?dydydy,,11.．设?，其中f(t)的三阶导数存在，且f??(t)?0，求 23?y?tf(t)?f(t)dxdxdx?dyy?(t)f?(t)?tf??(t)?f?(t)解：???t
dxx?(t)f??(t)
dy
2dxdydx332?ddxd2dxdx(dydx2(dy)?ddx(ddx1(t)?)?ddt1f??(t)(1) dt??f???(t)?1f??(t)??f???(t)?)?f??(t)f??(t)dx?f??(t)?2?f??(t)?3
12.．设方程xy2?ey?cos(x?y2),求y? 解：y2?2xyy??eyy???sin(x?y2)?(1?2yy?) y???
13．设有方程2x?tan(x?y)?解：方程两边对x求导，可得
2?sec2(x?y)?(1?y?)?sec2(x?y)?(1?y?)?1?y??1sec(x?y)2y?sin(x?y)2xy?e?2ysin(x?y)y222
?x?y0sectdt,求
2dydx22
?y??1?cos2(x?y)?sin2(x?y)?y???2sin(x?y)cos3(x?y) 14．设由方程xy?yx确定y是x的函数，求解：eylnx?exlny，两边同时x对求导，得e?x(y?lnx?ydydx
(y?lnx?yx)?exlnyylnx(lny?xy?x)
y??dyy(y?xlny)x? )?y?lny?x???xx?dxx(x?ylnx)?y(n)15.设f(x)任意阶可导，且f?(x)?e?f(x),f(0)?1,求f?f(x)?2f(x)解：f??(x)??e f?(x)??e?2f(x)?3f(x) f???(x)??2e f?(x)?2e(0)
f f(4)(x)??3?2e(x)?(?1)(n)6?3f(x)?4f(x) f?(x)??3?2e ?
(n)n?1(n?1)!en?16?nf(n)
?(?1)n?1所以，f(0)?(?1)23(n?1)!ex，求f23?nf(0)(n)(n?1)!e4?n
216．设y?sin解：y?(sin ?sin y(n)4x?cos
2x)?(cosx?sinn2x)?(sin2x?cos2x)(sin2x?sin2xcos34sin2x?cos2x?584x) 38cos4x
xcosx?cos4x??3sinxcosx?1?2??38?4?cos(4x?n??2)
17．设y?11?x2arcsinx,求y(n)(0)
解：y??x1?x2?11?x2arcsinx?11?x22?(1?x)y??xy?1?0
?(1?x2)y???3xy??y?0 ?(1?x2)y????5xy???4y??0 ?
?(1?x2)y(n?1)?(2n?1)xy(n)?n2y(n?1)?0 显然，当x?0时，y???1,y???0,y?????4,?? 故 y(2n)(0)?0,y(2n?1)(0)??4n(n!)2 18.求?原式=?1sin2x?2cosx2dx
d(tanx)2?1(sin22xxcos?2)cosx2dx=?tan2=
x12arctantanx2?c
19．求?1?lnx(x?lnx)1?lnx2dx
原式=?(xx?lnxx12121232dx=-?21)(x?lnxx121414d(2x?lnxx)=
xx?lnx+c
)20. 求不定积分?(x解：原式? ? ??2x?5)cos2xdx
(x?2x?5)sin2x?(x?2x?5)sin2x?(x?2x?5)sin2x?1?sinxedx
x33?(3x?2)sin2xdx
(3x?2)cos2x?(3x?2)cos2x?3323234?xcos2xdxxsin2x?38
cos2x?C
21．求不定积分?1?cosx1?sinxx11xxxedx??edx??tan?edx 解：原式??222x2x2coscos221xxxxxxxxed()??tan?edx?tan?e??tan?edx? ??22222xcos2xx ?tan?e?C
2?tanx2?edx
x
22．求不定积分?dxx(x10?1)
解：令x10?u，则du?10x9dx，于是
原式?1101?u(u?1)du2?110??11du??22??10?u(u?1)(u?1)u(u?1)u?1?u1??du ???du ? ? ??111??2??10?uu?1(u?1)1?1??lnu?lnu?1???C 10?u?1?1110 ?lnx?lnx?1??C 101010(x?1)23．求不定积分?解：??1sin3x2dx
13sinx
dx??sinx?cossin32xxdx??cscxdx??cot2x?cscxdx
?cscxdx??cotxd(cscx)?lncscx?cotx?cotxcscx??sin13?2cscx?(?cscx)dx ???lncscx?cotx?cotxcscx?xdx
故?1sin3xdx?12lncscx?cotx?sinxx12cotxcscx?C
24. 设f(x)的原函数为解：?xf?(2x)dx?12，求解不定积分?xf?(2x)dx
12xf(2x)?12?xd(f(2x))??f(2x)dx?12xf(2x)?14?f(2x)d(2x)
?xcosx?sinx?sinx?因为f(x)的原函数为，故f(x)?? ??2xxx??2xcos2x?sin2x于是，f(2x)?， 24x2xcos2x?sin2xsin2x1sin2x??C?cos2x??C 故?xf?(2x)dx?8x8x44xsinx25．设f(x)在?0,???上连续且满足?解：将方程?x(1?x)2x(1?x)20f(t)dt?x，求f(2)
0f(t)dt?x的两边对x求导，得
fx(1?x)?x(1?x)?2??2???1，即f(x2?x)?(x?x)?1
323令x?1,得f(2)?15