20xx陈文灯考研数学复习指南习题详解(理工)-高等数学ch1(编辑修改稿)

20xx陈文灯考研数学复习指南习题详解(理工)-高等数学ch1(编辑修改稿)内容摘要：

ns i nt a ns i n10 s in1s int a n1lim xxxxxxxx xxx     = 30 sintanlim x xxxe  = 30 )cos1(sinlim x xxxe  = 212sin2sinlim 3 20 ee x xxx  . 2. 求下列极限 (1) 3 231 12arc s in)11ln (lim  xxx 解 . 当 x1 时 , 33 1~)11ln (  xx , 3 23 2 12~12a r c s in  xx . 按照等价无穷小代换 3313 2313 231 22112 1lim12 1lim12a r c s i n )11l n(lim  xxxxxxxx (2)   xxx 220 cot1lim 解 . 方法 1：   xxx 220 cot1lim =   xxxx 2220 sinc os1lim =   xx xxxx 22 2220 s in c o ss inlim =   4220c o s)1(1lim x xxx=   3220 4s inc o s)1(2c o s2lim x xxxxxx =320320 4s i nco s2l i m4 2s i nco s2l i m x xxxx xxxxx   5 = 2112 2co s2s i nco s4co s2lim220  xxxxxxx = 213124 2s i n4s i nco s4lim213112 2co s2co s2lim0220  xxxxx xxxx = 32213161213124 2s in2lim0  x xx 方法 2：   xxx 220 cot1lim=   xxxx 2220 sinc os1lim=   xx xxxx 222220 s inc o ss inlim =   4220c o s)1(1lim x xxx=  420)12) ( c o s1(211lim xxxx =  444220)(0!4 )2(!2 )2(11)(1(211l i m xxxxxx =  4442420))(024162222(211l i m xxxxxxx = 3232lim440  xxx 3. 求下列极限 (1) )1(lnlim  nn nnn 解 . nnnnn nnnnn ln 1lim)1(lnlim  xnn 1令 1)1ln(lim0  xxx (2) nxnxn ee 11lim 解 .  10111limnxnxn ee 000xxx (3) nnnnba 。