Question

如图，求过程，谢谢

Answer 1

lim(x->0) [sin6x +xf(x)] /x^3 = 0 (0/0)

lim(x->0) [6cos6x +f(x)+xf'(x)] /(3x^2) = 0 (0/0)

=>
6+ f(0)=0
f(0)=-6 (1)

lim(x->0) [6cos6x +f(x)+xf'(x)] /(3x^2) = 0 (0/0)
lim(x->0) [-36sin6x +2f'(x)+xf''(x)] /(6x) = 0 (0/0)
=>
f'(0) =0 (2)

lim(x->0) [-36sin6x +2f'(x)+xf''(x)] /(6x) = 0 (0/0)
lim(x->0) [-216.cos6x +3f''(x)+xf'''(x)] /6 = 0
=>
-216 +3f''(0)+ =0
f''(0) =72

lim(x->0) [6+f(x)] /x^2 (0/0)
=lim(x->0) f'(x) /(2x) (0/0)
=lim(x->0) f''(x) /2
=72/2
=36
感觉提问主意不是很清晰
这里的只能参考了

追问

谢谢你

这不是我想要的