1.∫(0,x)f(x-t)dt=∫(x,0)f(u)d(x-u)
=∫(0,x)f(u)du=∫(0,x)f(t)dt
∴[∫(x-t)f(t)dt]/[x∫f(x-t)dt]=[x∫f(t)dt-∫tf(t)dt]/[x∫f(t)dt]
=1-∫tf(t)dt/[x∫f(t)dt]
∴lim[∫(x-t)f(t)dt]/[x∫f(x-t)dt]
=1-lim∫tf(t)dt/[x∫f(t)dt] 为0/0型 由洛必达法则有
lim∫tf(t)dt/[x∫f(t)dt]=lim xf(x)/[∫f(t)dt+xf(x)] 有由积分中值定理
原式=lim xf(x)/[xf(x0)+xf(x)] (x0在0,x之间)
=lim f(x)/[f(x0)+f(x)] 又f(0)≠0
∴limf(x)/[f(x0)+f(x)]=f(0)/2f(0)=1/2
∴原极限为1/2
2.(2n+1)*x^n=2(n+1)*x^n-x^n
=2[x^(n+1)]'-x^n
∴∑（2n+1)*x^n=2[∑x^(n+1)]'-∑x^n
=2{[x-x^(n+2)]/(1-x)}'-[1-x^(n+1)]/(1-x)
=2[1-(n+2)*x^(n+1)+(n+1)*x^(n+2)]/(1-x)^2-[1-x^(n+1)]/(1-x) (x≠1)
∑（2n+1)*x^n=n^2+n+n=n^2+2n (x=1)
3.应有x→1
分子分母同时趋近于0,则1+a+b=0
又sin(x^2-1)～(x^2-1)
∴lim(x^2+ax+b)/sin(x^2-1)
=lim(x^2+ax-1-a)/(x^2-1)
=lim(x+1+a)/(x+1)=(a+2)/2=3 得a=4,b=-5 1.∫(0,x)f(x-t)dt=∫(x,0)f(u)d(x-u)
=∫(0,x)f(u)du=∫(0,x)f(t)dt
∴[∫(x-t)f(t)dt]/[x∫f(x-t)dt]=[x∫f(t)dt-∫tf(t)dt]/[x∫f(t)dt]
=1-∫tf(t)dt/[x∫f(t)dt]
∴lim[∫(x-t)f(t)dt]/[x∫f(x-t)dt]
=1-lim∫tf(t)dt/[x∫f(t)dt] 为0/0型 由洛必达法则有
lim∫tf(t)dt/[x∫f(t)dt]=lim xf(x)/[∫f(t)dt+xf(x)] 有由积分中值定理
原式=lim xf(x)/[xf(x0)+xf(x)] (x0在0,x之间)
=lim f(x)/[f(x0)+f(x)] 又f(0)≠0
∴limf(x)/[f(x0)+f(x)]=f(0)/2f(0)=1/2
∴原极限为1/2
2.(2n+1)*x^n=2(n+1)*x^n-x^n
=2[x^(n+1)]'-x^n
∴∑（2n+1)*x^n=2[∑x^(n+1)]'-∑x^n
=2{[x-x^(n+2)]/(1-x)}'-[1-x^(n+1)]/(1-x)
=2[1-(n+2)*x^(n+1)+(n+1)*x^(n+2)]/(1-x)^2-[1-x^(n+1)]/(1-x) (x≠1)
∑（2n+1)*x^n=n^2+n+n=n^2+2n (x=1)
3.应有x→1
分子分母同时趋近于0,则1+a+b=0
又sin(x^2-1)～(x^2-1)
∴lim(x^2+ax+b)/sin(x^2-1)
=lim(x^2+ax-1-a)/(x^2-1)
=lim(x+1+a)/(x+1)=(a+2)/2=3 得a=4,b=-5