Question

极限问题。泰勒公式

Answer 1

解：∵(1+x)^α=1+αx+α(α-1)(x²/2)+o(x²)
(泰尺改勒公式，o(x)是高阶无穷小)
∴(x³+3x²)^(1/3)=x(1+3/x)^(1/誉困启3)
=x[1+(1/3)(3/x)+(1/3)(1/3-1)((3/x)²/2)+o(1/x²)]
(应用上式泰勒公式展开)
=x[1+1/x-1/x²+o(1/x²)]
=x+1-1/x+o(1/x)
(x^4-2x³)^(1/4)=x(1-2/x)^(1/4)
=x[1+(1/4)(-2/x)+(1/4)(1/4-1)((-2/x)²/2)+o(1/x²)]
(应用上式泰勒公庆如式展开)
=x[1-1/(2x)-3/(8x²)+o(1/x²)]
=x-1/2-3/(8x)+o(1/x)
故
原式=lim(x->∞)[(x+1-1/x+o(1/x))-(x-1/2-3/(8x)+o(1/x))]
=lim(x->∞)[3/2-5/(8x)+o(1/x)]
=3/2-0+0
(lim(x->∞)[o(1/x)]=0)
=3/2。