Question

问一道高数题 若x->0时，αxα与x2(ex-1）+2x4+3x5是同阶无穷小，则α=？

Answer 1

这题要用洛必达法则，铅余培洛必达之后，分母求导得a^2x^(a-1),分子求导得(2x+x^2)e^x-2x+8x^3+15x^4,约分得(2e^x+xe^x-2+8x^2+15x^3)/[a^2x^(a-2)],这时分子的极限为2，因此分母必须满足a-2=0，两个槐唯极限才能同阶，因毁和此a=2.

Answer 2

x->0
e^x = 1+x+(1/2)x^2 +(1/6)x^3 +o(x^3)
x^2.(e^x -1) =x^3+(1/芦派昌羡薯2)x^4 +(1/陪扒6)x^5 +o(x^5)
x^2.(e^x -1) +2x^4 +3x^5
=[x^3+(1/2)x^4 +(1/6)x^5 +o(x^5)] +2x^4 +3x^5
=x^3 +o(x^3)
=>α=3

追问

答案是α=3

追答

α=3, 一时手快！

追问

=x^3 +o(x^3) 这个是怎么得出的

追答

o(x^3) 代表所有的x^n 项 (n>3)

x^2.(e^x -1) +2x^4 +3x^5

=[x^3+(1/2)x^4 +(1/6)x^5 +o(x^5)] +2x^4 +3x^5

=x^3 + (3/2)x^4 +(19/6)x^5 +o(x^5)

=x^3 + [(3/2)x^4 +(19/6)x^5 +o(x^5)]

[(3/2)x^4 +(19/6)x^5 +o(x^5)] : 用o(x^3) 代替

=x^3 +o(x^3)