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limx→0[cos(sinx)-cosx]/x^4,sin(sinx)为什么不能换成sinx
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x->0
sinx ~ x - (1/6)x^3
cos(sinx)
~ 1 -(1/2)[x - (1/6)x^3]^2 + (1/24)[x - (1/6)x^3]^4
~ 1 - (1/2)[x^2 - (1/3)x^4] +(1/24)[ x^4]
~ 1 -(1/2)x^2 + (5/24)x^4
cosx ~1 -(1/2)x^2 +(1/24)x^4
cos(sinx) - cosx ~ (1/6)x^4
lim(x->0) [cos(sinx) -cosx ]/x^4
=lim(x->0) (1/6)x^4/x^4
=1/6
Or
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3) (0/0)
分母: x^3
分子: 也要到 x^3
sin(sinx)
~ sin ( x - (1/6)x^3)
~ [ x - (1/6)x^3] - (1/6)[ x - (1/6)x^3]^3
~ x - (1/6)x^3 -(1/6)x^3
~ x - (1/3)x^3
cosx ~ 1 - (1/2)x^2
cosx. sin(sinx)
~ [ 1 - (1/2)x^2] . [x - (1/3)x^3]
~ x - (1/3)x^3 - (1/2)x^3
~ x - (5/6)x^3
sinx ~ x- (1/6)x^3
-cosx. sin(sinx) + sinx
~-[x - (5/6)x^3] +[x- (1/6)x^3]
~ (2/3)x^3
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3)
=lim(x->0) (2/3)x^3/(4x^3)
=1/6
sinx ~ x - (1/6)x^3
cos(sinx)
~ 1 -(1/2)[x - (1/6)x^3]^2 + (1/24)[x - (1/6)x^3]^4
~ 1 - (1/2)[x^2 - (1/3)x^4] +(1/24)[ x^4]
~ 1 -(1/2)x^2 + (5/24)x^4
cosx ~1 -(1/2)x^2 +(1/24)x^4
cos(sinx) - cosx ~ (1/6)x^4
lim(x->0) [cos(sinx) -cosx ]/x^4
=lim(x->0) (1/6)x^4/x^4
=1/6
Or
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3) (0/0)
分母: x^3
分子: 也要到 x^3
sin(sinx)
~ sin ( x - (1/6)x^3)
~ [ x - (1/6)x^3] - (1/6)[ x - (1/6)x^3]^3
~ x - (1/6)x^3 -(1/6)x^3
~ x - (1/3)x^3
cosx ~ 1 - (1/2)x^2
cosx. sin(sinx)
~ [ 1 - (1/2)x^2] . [x - (1/3)x^3]
~ x - (1/3)x^3 - (1/2)x^3
~ x - (5/6)x^3
sinx ~ x- (1/6)x^3
-cosx. sin(sinx) + sinx
~-[x - (5/6)x^3] +[x- (1/6)x^3]
~ (2/3)x^3
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3)
=lim(x->0) (2/3)x^3/(4x^3)
=1/6
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追问
没有解释为什么sin(sinx)不能直接等价于sinx
追答
lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3) (0/0)
分母 : order x^3
分子 : 一定要到 x^3
根据Taylor's expansion of sinx
sinx = x -(1/6)x^3 + (1/120)x^5+....
sin(sinx)等价于sinx : 这个明显不足,一定要
sin(sinx) ~ sin[x -(1/6)x^3]
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