微積分問題 請幫幫忙

1.
f(x)=(x-3)/[(x-2)(x+1)]=(1/3)[ 4/(x+1)- 1/(x-2)]
=(4/3)* 1/(1+x) + (1/6) * 1/[1- (x/2)]
=(4/3) [1-x+x^2-...+x^100- x^101+...] + (1/6)[ 1+(x/2)+(x/2)^2+...+(x/2)^100+...]
= ...+[(4/3)+(1/6)/2^100] x^100+...
= ...+f^(100)(0)/100! * x^100+ ...
故 f^(100)(0)= 100!*[(4/3)+(1/6)/2^100]

2.
∫[0~1] f^(-1)(y) dy (令 y=f(x))
=∫[0~1] x*f'(x) dx (integration by parts)
= [xf(x)]_[0~1] - ∫[0~1] f(x) dx
= f(1)-0- ∫[0~1] f(x)dx
= 1- (2/3) = 1/3

3.
設題目是 lim(x->0) [sin(x^2)-(sinx)^2]/(arcsin x )^4] (???)
Maclaurin's series: sinx=x- x^3/3!+...
so, sin(x^2)= x^2- x^6/6+... ----(A)
(sinx)^2=(x- x^3/6+...)^2= x^2 - x^4/3+ ... ---(B)
又1/√(1-x)= (1-x)^(-1/2) = 1+(1/2)x- (3/8)x^2+...
so 1/√(1-x^2)= 1+(1/2)x^2 -(3/8)x^4+...
積分得 arcsin(x)= x-(1/6)x^3-(3/40)x^5+...
4次方得 [arcsin(x)]^4= x^4 -(4/6)x^6+... ----(C)
將(A),(B),(C)代入所求limit得
原式= lim(x->0) [(1/3)x^4+...]/[x^4-(4/6)x^6+...] = 1/3

第三題我有不同解法
只要把題目稍微整理成 幾個式子就OK了

(sinx^2-x^2+x^2-sin^2(x)) / {(sin^(-1)(x))^4 }

=> [ (x^2)( (sinx^2-x^2)/x^6 ) +( (x+sinx) /x )( (x-sinx) /x^3 ) ] / (sin^(-1)(x))^4/x^4)


寫到這就可以得到答案了
已知 lim x→0 (x-sinx)/x^3 = 1/6 lim x→0 arcsinx/x =1


2011-05-13 22:47:17 補充：
小弟我應該沒打錯符號吧 字太多容易打錯 ^^

第三題我忘了利用麥克勞林
高手果真是高手><"