Question

请问这几道高数题的极限怎么算？

Answer 1

主要是运用了等价无穷小公式，如下详细的公式和详解望采纳

Answer 2

(1)
lim(x->0) sin(x^n)/ (sinx)^m
=lim(x->0) x^n/ x^m
=lim(x->0) x^(n-m)
=0 ; n>m
=1 ; n=m
不存在 ; n<m
(2)
lim(x->∞) [(x-1)/x]^[1/sin(1/x)]
=lim(x->∞) (1 - 1/x )^[1/sin(1/x)]
=lim(x->∞) (1 - 1/x )^x
=e^(-1)
(3)
lim(x->1) arcsin(1-x)/ lnx
=lim(x->1) arcsin(1-x)/ ln[1+(x-1)]
=lim(x->1) (1-x)/ (x-1)
=-1
(4)
x->0
分子
1-cosx = (1/2)x^2+o(x^2)
分母
1-cos√x = (1/2)x +o(x)
x(1-cos√x) = (1/2)x^2 +o(x^2)
lim(x->0) [ 1- √cosx]/[x(1-cos√x)]

=lim(x->0) ( 1- cosx)/{ [x(1-cos√x)] .[ 1+ √cosx] }
=(1/2)lim(x->0) ( 1- cosx)/{ [x(1-cos√x)]

=(1/2)lim(x->0) (1/2)x^2/[ (1/2)x^2 ]
=1/2