Question

求下列不定式极限：

Answer 1

很简单的，就是步骤烦了一些。注意，在运用洛必达之前，要先看看能不能等阶替换，还有看看能不能化简，不然繁死人.

Answer 2

(1)
lim(x->0) [ 1/(1-cosx) - 2(cotx)^2 ]
=lim(x->0) [ 1/(1-cosx) - 2(cosx)^2/(sinx)^2 ]
=lim(x->0) [ (sinx)^2 - 2(1-cosx)(cosx)^2 ]/[(1-cosx)(sinx)^2 ]
=lim(x->0) [ (sinx)^2 - 2(1-cosx)(cosx)^2 ]/[ (1/2)x^4 ]
=lim(x->0) [ (sinx)^2 - 2(cosx)^2 + 2(cosx)^3 ]/[ (1/2)x^4 ]
=lim(x->0) [ 1 - 3(cosx)^2 + 2(cosx)^3 ]/[ (1/2)x^4 ]
(0/0 分子分母分别求导)
=lim(x->0) [ 6cosx.sinx - 6(cosx)^2. sinx ]/( 2x^3 )
=lim(x->0) (3sin2x - 3sin2x.cosx)/( 2x^3 )
=[lim(x->0) 3sin2x/x ] . lim(x->0) (1 - cosx)/( 2x^2 )
=6lim(x->0) (1 - cosx)/( 2x^2 )
=6lim(x->0) (1/2)x^2/( 2x^2 )
=6/4
=3/2

(2)

x->0
ln(1+x) = x -(1/2)x^2 +o(x^2)
e^x = 1+ x+(1/2)x^2 +o(x^2)
ln(1+x) -e^x +1 = -x^2 +o(x^2)
lim(x->0) [ 1/(e^x-1) - 1/ln(1+x) ]
=lim(x->0) [ln(1+x) - e^x +1 ]/[(e^x-1).ln(1+x) ]
=lim(x->0) [ln(1+x) - e^x +1 ]/x^2
=lim(x->0) -x^2/x^2
=-1