Question

数学求极限，要有过程

Answer 1

取数用等价穷洛必达则
lim
ln(sinx/x)/(1-cosx)
=lim
(sinx/x-1)/(1-cosx)
=lim
(sinx-x)/x(1-cosx)
=lim
(sinx-x)/(1/2*x^3)
=lim
(cosx-1)/(3/2*x^2)
=lim
(-1/2*x^2)/(3/2*x^2)
=-1/3
所原极限e^(-1/3)
这个仅供参考，你可以看下解题思路，如果不对请忽略。