第1个回答 2021-07-01
如图所示：
第2个回答 2021-07-01
x->0
sin6x =6x -(1/6)(6x)^3 + o(x^3) = 6x-36x^3 +o(x^3)
xf(x)
= x [ f(0) + f'(0)x + (1/2)f''(0)x^2 +o(x^2) ]
=xf(0) + f'(0)x^2 +(1/2)f''(0)x^3 +o(x^3)
sin6x +xf(x)
=[6-f(0)]x + f'(0)x^2 +[-36 +(1/2)f''(0) ]x^3 +o(x^3)
lim(x->0) [sin6x +xf(x)] /x^3 =0
=>
6-f(0)=0 and f'(0)=0 and -36 +(1/2)f''(0)=0
f(0) =6 and f'(0)=0 and f''(0) =72
lim(x->0) (6+f(x))/x^2
洛必达
=lim(x->0) f'(x)/(2x)
洛必达
=lim(x->0) f''(x)/2
=f''(0)/2
=72/2
=36
sin6x =6x -(1/6)(6x)^3 + o(x^3) = 6x-36x^3 +o(x^3)
xf(x)
= x [ f(0) + f'(0)x + (1/2)f''(0)x^2 +o(x^2) ]
=xf(0) + f'(0)x^2 +(1/2)f''(0)x^3 +o(x^3)
sin6x +xf(x)
=[6-f(0)]x + f'(0)x^2 +[-36 +(1/2)f''(0) ]x^3 +o(x^3)
lim(x->0) [sin6x +xf(x)] /x^3 =0
=>
6-f(0)=0 and f'(0)=0 and -36 +(1/2)f''(0)=0
f(0) =6 and f'(0)=0 and f''(0) =72
lim(x->0) (6+f(x))/x^2
洛必达
=lim(x->0) f'(x)/(2x)
洛必达
=lim(x->0) f''(x)/2
=f''(0)/2
=72/2
=36
第3个回答 2021-07-01
关于大学电路的一些问题
关于大学电路的一些问题刚学电路有点懵，元件放出或者吸收功率必须要带正负号吗？可不可以把功率都写成正的直接根据关不关联来判断？