论高等数学的极限

溶液1

利用辛克斯的麦克劳克林公式展开

sin6x=6x-(6x)^3/3！+o(x^3)

F(x)展开f(x)= f(0)+f '(0)x+1/2f ' '(0)x2+o(x2)

替代获得

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

lim[6x+f(0)x+f '(0)x 2]/x 3+1/2f ' '(0)-36 = 0。

所以f(0)=-6 f '(0)= 0 1/2 f ' '(0)-36 = 0 f ' '(0)= 72。

林[6+ f(x)]/x^2=limf''(0)/2=36

解决方案2

lim[6+f(x)]/x^2=lim[6+f(x)]/x^2-0

=lim[6+f(x)]/x^2-lim[sin6x+xf(x)]/x^3

=lim{[6+f(x)]/x^2-[sin6x+xf(x)]/x^3}

= lim [(6x-sin6x)/x 3](洛必达定律)

=lim[(6-6cos6x)/3x^2]

= 2li m [(1-cos6x)/x 2](等价无穷小代换)

=2lim[(1/2)(6x)^2/x^2]

=6^2=36