Question

求8.9两道题的极限

Answer 1

(8)
x^2+1
=x(x+1) - x+1
=x(x+1) - (x+1) +2
lim(x->+∞) [ (x^2+1)/(x+1)- ax-b ] = 1
lim(x->+∞) [ x -1 + 2/(x+1)- ax-b ] = 1
lim(x->+∞) [ (1-a)x +(-1-b) + 2/(x+1) ] = 1
=>
1-a =0 and -1-b=1
a=1 and b=-2
(9)

(1+2+...+n)/(n^2+n)≤[ 1/(n^2+1) + 2/(n^2+2)+...+n/(n^2+n) ]≤(1+2+...+n)/(n^2+1)

lim(n->+∞) (1+2+...+n)/(n^2+1)
=lim(n->+∞) n(n+2)/[2(n^2+1)]
=1/2
lim(n->+∞) (1+2+...+n)/(n^2+n)
=lim(n->+∞) n(n+1)/[2(n^2+n)]
=1/2
=>
lim(n->+∞) [ 1/(n^2+1) + 2/(n^2+2)+...+n/(n^2+n) ] =1/2

Answer 2

8、lim(x->∞) [(x^2+1)/(x+1)-ax-b]
=lim(x->∞) (x^2+1-ax^2-bx-ax-b)/(x+1)
=lim(x->∞) [(1-a)x^2-(a+b)x+1-b]/(x+1)
=lim(x->∞) [(1-a)x-(a+b)+(1-b)/x]/(1+1/x)
=lim(x->∞) [(1-a)x-(a+b)]
=1
所以1-a=0，且a+b=-1
得a=1，b=-2
9、因为n^2+1<n^2+2<...<n^2+n
所以1/(n^2+n)+2/(n^2+n)+...+n/(n^2+n)
<1/(n^2+1)+2/(n^2+2)+...+n/(n^2+n)
<1/(n^2+1)+2/(n^2+1)+...+n/(n^2+1)
即1/2<1/(n^2+1)+2/(n^2+2)+...+n/(n^2+n)<(n^2+n)/2(n^2+1)
又因为lim(n->∞) (n^2+n)/2(n^2+1)
=lim(n->∞) (1+1/n)/2(1+1/n^2)
=1/2
所以根据极限的夹逼性，
lim(n->∞) 1/(n^2+1)+2/(n^2+2)+...+n/(n^2+n)=1/2