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lim<x→+∞>[(x^2+1)/(x+1) - ax + b]
= lim<x→+∞>[(1-a)x^2+(b-a)x+b+1]/(x+1) = 1/2
1-a = 0, b-a = 1/2, 解得 a = 1, b = 3/2, a＋b ＝ 5/2 .
= lim<x→+∞>[(1-a)x^2+(b-a)x+b+1]/(x+1) = 1/2
1-a = 0, b-a = 1/2, 解得 a = 1, b = 3/2, a＋b ＝ 5/2 .
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x^2+1
=x(x+1) -x+1
=x(x+1) -(x+1) +2
(x^2+1)/(x+1) = x -1 +2/(x+1)
//
lim(x->+无穷) [(x^2+1)/(x+1) -ax+b ] =1/2
lim(x->+无穷) [ x -1 +2/(x+1) -ax+b ] =1/2
lim(x->+无穷) [ (1-a)x +(-1+b) +2/(x+1) ] =1/2
=>
1-a=0 and -1+b=1/2
a=1 and b=3/2
a+b = 1+3/2 = 5/2
=x(x+1) -x+1
=x(x+1) -(x+1) +2
(x^2+1)/(x+1) = x -1 +2/(x+1)
//
lim(x->+无穷) [(x^2+1)/(x+1) -ax+b ] =1/2
lim(x->+无穷) [ x -1 +2/(x+1) -ax+b ] =1/2
lim(x->+无穷) [ (1-a)x +(-1+b) +2/(x+1) ] =1/2
=>
1-a=0 and -1+b=1/2
a=1 and b=3/2
a+b = 1+3/2 = 5/2
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