数学 画圈的两题？
(1)
x->0
sin4x = 4x -(1/6)(4x)^3 +o(x^3) = 4x - (32/3)x^3 +o(x^3)
(1/2)sin4x = 2x - (16/3)x^3 +o(x^3)
2x-(1/2)sin4x = (16/3)x^3 +o(x^3)
//
lim(x->0) [ 1/(sinx)^2 - (cosx)^2/x^2]
=lim(x->0) [x^2- (sinx.cosx)^2 ]/[(sinx)^2.x^2]
=lim(x->0) [x^2- (sinx.cosx)^2 ]/x^4
=lim(x->0) [x^2- (1/4)(sin(2x))^2 ]/x^4
洛必达
=lim(x->0) ( 2x- (1/2)sin4x )/(4x^3)
=lim(x->0) (16/3)x^3/(4x^3)
=4/3
(2)
x->0
ln(1+x^2)= x^2 -(1/2)x^4 +o(x^4)
tanx = x+(1/3)x^3 +o(x^3)
(tanx)^2 =[x+(1/3)x^3 +o(x^3)]^2 = x^2 +(2/3)x^4 +o(x^4)
ln(1+(tanx)^2)
=ln[1+ x^2 +(2/3)x^4 +o(x^4)]
=[x^2 +(2/3)x^4 +o(x^4)] -(1/2)[ x^2 +(2/3)x^4 +o(x^4)]^2 +o(x^4)
=[x^2 +(2/3)x^4 +o(x^4)] -(1/2)[ x^4 +o(x^4)] +o(x^4)
=x^2 +(2/3-1/2)x^4 +o(x^4)
=x^2 +(1/6)x^4 +o(x^4)
ln(1+x^2) -ln(1+(tanx)^2)
=[x^2 -(1/2)x^4 +o(x^4)] -[x^2 +(1/6)x^4 +o(x^4)]
=-(2/3)x^4+o(x^4)
//
lim(x->0) [ 1/ln(1+(tanx)^2) - 1/ln(1+x^2) ]
=lim(x->0) [ ln(1+x^2) -ln(1+(tanx)^2) ]/[ln(1+(tanx)^2).ln(1+x^2) ]
=lim(x->0) [ ln(1+x^2) -ln(1+(tanx)^2) ]/x^4
=lim(x->0) -(2/3)x^4/x^4
=-2/3
不用泰勒直接做可以吗？
不用泰勒，即是用洛必达，那计算上比较麻烦！
lim(x->0) [ 1/(sinx)^2 - (cosx)^2/x^2]
=lim(x->0) [x^2- (sinx.cosx)^2 ]/[(sinx)^2.x^2]
=lim(x->0) [x^2- (sinx.cosx)^2 ]/x^4
=lim(x->0) [x^2- (1/4)(sin(2x))^2 ]/x^4
洛必达
=lim(x->0) ( 2x- (1/2)sin4x )/(4x^3)
洛必达
=lim(x->0) ( 2- 2cos4x )/(12x^2)
=lim(x->0) 2( 1- cos4x )/(12x^2)
=lim(x->0) 16x^2/(12x^2)
=4/3
(2)
lim(x->0) [ 1/ln(1+(tanx)^2) - 1/ln(1+x^2) ]
=lim(x->0) [ ln(1+x^2) -ln(1+(tanx)^2) ]/[ln(1+(tanx)^2).ln(1+x^2) ]
=lim(x->0) [ ln(1+x^2) -ln(1+(tanx)^2) ]/x^4
洛必达
=lim(x->0) [ 2x/(1+x^2) -2tanx.(secx)^2/(1+(tanx)^2) ]/(4x^3)
再计算下去比较麻烦
。。。
=-2/3
