Question

求极限lim(x→0)(1/ln(1+x2)-1/(sinx)2)

Answer 1

简单计算一下即可，答案如图所示

Answer 2

lim(x->0) [ 1/ln(1+x^2) -1/(sinx)^2 ]
=lim(x->0) [ (sinx)^2 - ln(1+x^2) ]/[ (sinx)^2. ln(1+x^2) ]
x->0
分子
sinx~ x -(1/6)x^3
(sinx)^2
~ ( x- (1/6)x^3)^2
~ x^2 - (1/3)x^4
ln(1+x^2) ~ x^2 - (1/2)x^4
(sinx)^2 - ln(1+x^2) ~ (1/6)x^4
分母

(sinx)^2. ln(1+x^2) ~ x^4
lim(x->0) [ 1/ln(1+x^2) -1/(sinx)^2 ]
=lim(x->0) [ (sinx)^2 - ln(1+x^2) ]/[ (sinx)^2. ln(1+x^2) ]
=lim(x->0) (1/6)x^4/x^4
=1/6

Answer 3

你看下用的泰勒

Answer 4

Answer 5

=lim(sin²x-ln(1+x²))/x^4
=lim(2sinxcosx-2x/(1+x²))/4x³
=lim(sin2x(1+x²)-2x)/4x³
=lim(2cos2x(1+x²)+2xsin2x-2)/12x²
=lim(-4sin2x(1+x²)+4xcos2x+2sin2x+4xcos2x)/24x
=(-4*2+4+2*2+4)/24
=1/6