Question

高数问题 请问极限里当x趋近于零时，√cosx-1怎么等于1/2(cosx-1)？

第二行的cosx那块怎么到第三行的？求大神赐教

Answer 1

方法如下，
请作参考：

Answer 2

简单计算一下即可，答案如图所示

Answer 3

可以用二项展开：
√cosx -1 = [1+(cosx-1)]^(1/2) -1 ~ 1 + (1/2)(cosx-1) - 1 = (1/2)(cosx-1)

Answer 4

x->0
cosx = 1-(1/2)x^2 +o(x^2)
√cosx =[1-(1/2)x^2 +o(x^2)]^(1/2) = 1-(1/4)x^2 +o(x^2)
(1+(sinx)^2) = 1+ x^2+o(x^2)
(1+(sinx)^2)^(1/3) =[1+ x^2+o(x^2)]^(1/3) =1+(1/3)x^2 +o(x^2)
√cosx - (1+(sinx)^2)^(1/3)
=[ 1-(1/4)x^2 +o(x^2)] -[1+(1/3)x^2 +o(x^2)]
=-(7/12)x^2 +o(x^2)
lim(x->0+) [√cosx - (1+(sinx)^2)^(1/3) ]/x^2
=lim(x->0+) -(7/12)x^2/x^2
=-7/12