一道极限题lim(x→0)√(tansinx)/(sintanx)

题目描述貌似有些不清,再叙述一下lim(x→0)√(tansinx/sintanx)

先不管根号
就看
tansinx/sintanx
当x->0时 是 0/0的形式
所以应用罗必塔法则,即对分子分母分别求导得
(tansinx)'=1/cos^2(sinx) * cosx=cosx/cos^2(sinx)
(sintanx)'=costanx*1/cos^2x=costanx/cos^2x

所以
(tansinx)'/(sintanx)'=cosx/cos^2(sinx) / costanx/cos^2x = cos^3x / costanx * cos^2(sinx)
所以
lim(x->0)√(tansinx)/(sintanx)=lim(x->0)√cos^3x / costanx * cos^2(sinx)
=√1/cos(0)*cos^2(0)
=√1/1
=1
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第1个回答  2012-11-30
x→0
sinx/x→1
sintanx/tanx→1

lim(x→0)√(tansinx/sintanx)
=lim(x→0)√(tanx/tanx)
=1
第2个回答  2012-11-25
这是非而磁奖得主也做不出来
第3个回答  2012-11-25
我当初就是被数学整倒的,看见题鸡皮都起了

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