求助一道求极限问题,请问图里这个做法的问题出在哪里呢?
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发布时间：2024-10-11 07:57
共1个回答
热心网友
时间：2024-10-30 03:42
=>
sin6x - sinx. f(x) 的阶数 >3
x->0
sin6x = 6x -(1/6)(6x)^3 +o(x^3) = 6x -36x^3 +o(x^3)
sinx = x -(1/6)x^3 +o(x^3)
f(x) = f(0) +f'(0)x + (1/2)f''(0)x^2 +o(x^2)
sinx.f(x)
=[x -(1/6)x^3 +o(x^4)].[f(0) +f'(0)x + (1/2)f''(0)x^2 +o(x^2)]
=f(0)x +f'(0)x^2 + [ (1/2)f''(0)- (1/6)f(0)]x^3 +o(x^3)
sin6x -sinx.f(x)
=[6-f(0)]x +f'(0)x^2 +[-36 -(1/2)f''(0)+ (1/6)f(0)]x^3 +o(x^3)
阶数 >3
=>
6-f(0)=0 and f'(0)=0 and -36 -(1/2)f''(0)+ (1/6)f(0)=0
f(0) =6 and f'(0)=0 and f''(0) = -70
//
lim(x->0) [6-f(x)]/x^2
=lim(x->0) [6-f(0) -f'(0)x -(1/2)f''(0)x^2 ]/x^2
=lim(x->0) 35x^2/x^2
=35
