Question

高数如图

第一题的答案是2/3，两位都有疏忽，期待有简明完善的过程。满意会加。

Answer 1

Answer 2

(1)
lim(x->0)( 1/x^2-1/(tanx)^2)
=lim(x->0)( (tanx)^2-x^2) /[x^2(tanx)^2] (0/0)
=lim(x->0)[ (tanx)(secx)^2-x ] /[x^2(tanx)(secx)^2 + x(tanx)^2]
=lim(x->0)[ (tanx)((tanx)^2+1)-x ] /[x^2(tanx)(secx)^2 + x(tanx)^2]
=lim(x->0) (tanx)^3 /[x^2(tanx)(secx)^2 + x(tanx)^2] ( x~ tanx )
=lim(x->0) 1 /[(secx)^2 + 1]
= 1/2

(2)
f(π)=1
∫(0->π)[ f(x) + f''(x) ] sinx dx =3
f''(x) cts,
∫(0->π)[ f(x) + f''(x) ] sinx dx
=∫(0->π) f(x)sinxdx + ∫(0->π)f''(x)sinx dx
=-∫(0->π) f(x)dcosx + ∫(0->π)f''(x)sinx dx
=-[f(x)cosx](0->π) +∫(0->π) f'(x)cosxdx + ∫(0->π)f''(x)sinx dx
=f(π)+f(0) +∫(0->π) f'(x)dsinx + ∫(0->π)f''(x)sinx dx
=1+f(0) +[ f'(x)sinx](0->π) -∫(0->π) f''(x)sinxdx + ∫(0->π)f''(x)sinx dx
=1+f(0)
=> f(0) = 2

Answer 3

1 原式=lim{（tan^2 x-x^2）/x^4}=lim{（2tanxsec^2 x-2x）/4x^3}=lim{（2tanx-2xcos^2 x）/4x^3}=lim{（2sec^2 x-2cos^2 x+2xsin2x）/12x^2}=lim{（2(1-cos x)(1+COSX)(1+cos
^2 x)+2xsin2xcos^2x）/12x^2}=4/12+4/12=2/3
2 S（0，π）f（x）sinxdx+S（0，π）sinxdf‘（x）=S（0，π）f（x）sinxdx-S（0，π）f‘（x）cosxdx=S（0，π）f（x）sinxdx-f（x）cosx|（0，π）-S（0，π）f（x）sinxdx=-f（x）cosx|（0，π）=f（0）+f（π）=3
f（0）=2