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)

Question

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)

Answer 1

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