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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/21 06:22:18

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

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→0时：
3f''(x)+xf'''(x)=216
3f''(x)=216
f''(x)=72
所以：(x→0)lim[6+f(x)]/x^2
=(x→0)lim[f'(x)/(2x)]
=(x→0)lim[f''(x)/2]
=72/2
=36

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