Question

已知，lim x→0 [sin6x+xf(x)]/x^3=0 求：lim x→0 [6+f(x)]/x^2=?

我需要不用泰勒公式的解法！

Answer 1

lim(x-->0) [sin(6x) + xf(x)]/x³ = 0，要弄做不定式，分子和分母都要趋向0
sin(6x) + xf(x) = 0
xf(x) = - sin(6x)
f(x) = - (sin6x)/x

lim(x-->0) [6 + f(x)]/x²
= lim(x-->0) [6 - (sin6x)/x]/x²
= lim(x-->0) [6x - sin(6x)]/x³
= lim(x-->0) [6 - 6cos(6x)]/(3x²) <== 洛必达法则
= lim(x-->0) [- 6(- 6)sin(6x)]/(6x) <== 再洛必达法则
= lim(x-->0) 36 · sin(6x)/(6x)
= 36