Question

一道求极限的问题 lim[sin6x+xf(x)]/x^3=0,则lim[6+f(x)]/x^2= (x->0) (x->0)

Answer 1

利用sinx的麦克劳林公式展开
sin6x=6x-(6x)^3/3!+o(x^3）
f(x)在x=0处展开f(x)=f(0)+f'(0)x+1/2f''(0)x^2+o(x^2)
代入得到
lim[sin6x+xf(x)]/x^3=6x-(6x)^3/3!+o(x^3)+f(0)x+f'(0)x^2+1/2f''(0)x^3+o(x^3)/x^3=0 x→0
整理得lim[6x+f(0)x+f'(0)x^2]/x^3+1/2f''(0)-36=0
从而f(0)=-6 f'(0)=0 1/2f''(0)-36=0 f''(0)=72
lim[6+ f(x)]/x^2=limf''(0)/2=36
x→0
祝你学习愉快