Question

求解第二题，详细过程

高等数学 大学数学

Answer 1

f(0-)

=lim(x->0-) ln(1+ax^3)/(x-arcsinx)
=lim(x->0-) ax^3/[-(1/6)x^3]
=-6a
f(0) =6
f(0+)
=lim(x->0+) [e^(ax) +x^2-ax -1]/xsin(x/4)
=lim(x->0+) [e^(ax) +x^2-ax -1]/[ (1/4)x^2]
=lim(x->0+) [1 +(1/2)a^2]x^2/[ (1/4)x^2]
=4[1 +(1/2)a^2]
f(0+)=f(0)
4[1 +(1/2)a^2] =6
1 +(1/2)a^2 = 3/2
a^2 = 1
a=1 or -1
f(0-)=f(0)
-6a =6
a=-1
ie
a=-1
x=0 , f(x) 连续
//
x->0
e^(ax) = 1+ax +(1/2)a^2.x^2 +o(x^2)
e^(ax) +x^2-ax -1 = [1 +(1/2)a^2]x^2 +o(x^2)

追问

答案没