Question

若limx→0(sinx+xf(x))/x³=0则limx→0(1+f(x))/x²=？

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∵lim(x→0)
(sinx+xf(x))/x³=0.....(1)
且lim(x→0)
x³=0
∴lim(x→0)
(sinx+xf(x))=0
0/0型极限，用洛必达法则对分子分母同时求导
lim(x→0)
(sinx+xf(x))/x³=lim(x→0)
(cosx+f(x)+xf'(x))/3x²=0.....(2)
∵lim(x→0)
3x²=0
∴lim(x→0)
(cosx+f(x)+xf'(x))=0
即
1+lim(x→0)f(x)+0=0
∴lim(x→0)
f(x)=-1
对(2)继续用洛必达法则
lim(x→0)
(cosx+f(x)+xf'(x))/3x²=lim(x→0)
(-sin(x)+2f'(x)+xf''(x))/6x......(3)
同理可得
lim(x→0)
(-sin(x)+2f'(x)+xf''(x))=0
即
0+lim(x→0)
2f'(x)+0=0
∴lim(x→0)
f'(x)=0
对(3)继续用洛必达法则
lim(x→0)
(-sin(x)+2f'(x)+xf''(x))/6x=lim(x→0)
(-cos(x)+3f''(x)+xf'''(x))/6
同理可得
lim(x→0)
(-cos(x)+3f''(x)+xf'''(x))=0
即
-1+3lim(x→0)
f''(x)+0=0
∴lim(x→0)
f''(x)=1/3
lim(x→0)
(1+f(x))/x²
=lim(x→0)
f'(x))/2x......(∵lim(x→0)
f(x)=-1,∴lim(x→0)
1+f(x)=0,
0/0型,
用洛必达法则)
=lim(x→0)
f''(x))/2......(∵lim(x→0)
f'(x)=0,∴0/0型,
用洛必达法则)
=1/6