若limx趋近于0，（sin6x+xf（x））/x^3=0，求limx趋近于0，（6+f（x））/x^2的值

微积分的题求解

lim x→0,[sin6x + xf(x)]/x³=0+α，其中lim x→0,α=0
即f(x)/x² = -sin6x/x³ + α
从而lim x→0,[6+f(x)]/x²
=lim x→0,( 6/x² - sin6x/x³ + α )
=lim x→0,(6x-sin6x)/x³，用洛必达法则
=lim x→0,[6(1-cos6x)]/3x²，用等价无穷小lim x→0,(1-cosx)等价于lim x→0,x²/2
=lim x→0,[ 6 × (6x)² × 1/2 ]/3x²
=36

∵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