若lim [sin6x+xf(x)]/x^3=0,则lim [6+f(x)]/x^2是多少?（x是趋近0）可答案是36

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

答：(x→0)lim[sin6x+xf(x)]/x^3=0属于0-0型,可以应用洛必答法则：(x→0)lim[6cos6x+f(x)+xf'(x)]/(3x^2)=0(x→0)lim[-36sin6x+f'(x)+f'(x)+xf''(x)]/(6x)=0(x→0)lim[-216cos6x+2f''(x)+f''(x)+xf'''(x)]/6=0所以,x...