Question

求极限的题

Answer 1

1.此极限求出结果等于4/3。

2.求此题极限的过程见上图。

3.对于这1题，求极限时，是先通分，然后，分母利用等价无穷小代替的求极限的方法，再用洛必达法则的求极限的方法。最后一步，分母用等价无穷小代替的极限方法，就可以将极限求出了。

详细的求此题极限的步骤及说明见上。

Answer 2

(1)

x->0
sin(4x) =4x -(1/6)(4x)^3 +o(x^3) = 4x -(32/3)x^3 +o(x^3)
(1/2)sin(4x) = 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]/[x^2.(sinx)^2]
=lim(x->0) [x^2 - (sinx.cosx)^2]/x^4
=lim(x->0) [x^2 - (1/4)(sin2x)^2]/x^4
洛必达
=lim(x->0) [2x - (sin2x)(cos2x)]/(4x^3)
=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

Answer 3

4/3

应用知识点:

1:等价。

2:洛必达法则。

当x趋向于0

sinx~x

sin4x~4x

欢迎你追问。