Question

求极限当x→0若lim[sin6x+x f(x)]/x3=0,求lim[6+ f(x)]/x2

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