Question

求a，b的值？

Answer 1

x->0
sinx = x-(1/6)x^3 +o(x^3)
bx-sinx = (b-1)x +(1/6)x^3 +o(x^3)
lim(x->0) ∫(0->x) t^2/√(a+t) dt / ( bx-sinx) =1
=>
b-1 = 0
b=1
lim(x->0) ∫(0->x) t^2/√(a+t) dt / ( x-sinx) =1
lim(x->0) ∫(0->x) t^2/√(a+t) dt /[ (1/6)x^3] =1
lim(x->0) ∫(0->x) t^2/√(a+t) dt /x^3 =1/6
(0/0 分子分母分别求导)
lim(x->0) [ x^2/√(a+x) ] /(3x^2) =1/6
lim(x->0) [ 1/√(a+x) ] =1/2
1/√a =1/2
a=4
(a,b) = (4, 1)

追问

用洛必达法则为什么不行？

追答

行，为什么不行！
lim(x->0) ∫(0->x) t^2/√(a+t) dt / ( bx-sinx) =1
(0/0 分子分母分别求导)
lim(x->0) [ x^2/√(a+x)] / ( b-cosx) =1 (0/0)
=> b=1
lim(x->0) [ x^2/√(a+x)] / ( b-cosx) =1
lim(x->0) [ x^2/√(a+x)] / ( 1-cosx) =1
lim(x->0) [ x^2/√(a+x)] / [( 1/2)x^2] =1
lim(x->0) [ 1/√(a+x)] =1/2
1/√a =1/2
a=4

追问

感谢大佬，我终于知道我哪里不对了！