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(1)
x^2+x+1
=x(x+1) +1
lim(x->∞) [ (x^2+x+1)/(x+1) - ax -b)=0
lim(x->∞) [ (x+1) +1/(x+1) - ax -b)=0
lim(x->∞) [ (1-a)x+(1-b) +1/(x+1) ]=0
1-a=0 and 1-b=0
a=1 and b=1
(a,b) =(1,1)
(2)
lim(x->+∞) (x-1)(x-2)(x-3)(x-4)(x-5)/(4x-1)^a= b
分子: x^6 的系数 = 1
分母 : x^a 的系数 = 4^a
b≠0 => a=6
b= 1/4^6 =1/4096
(a,b) =(6, 1/4096)
x^2+x+1
=x(x+1) +1
lim(x->∞) [ (x^2+x+1)/(x+1) - ax -b)=0
lim(x->∞) [ (x+1) +1/(x+1) - ax -b)=0
lim(x->∞) [ (1-a)x+(1-b) +1/(x+1) ]=0
1-a=0 and 1-b=0
a=1 and b=1
(a,b) =(1,1)
(2)
lim(x->+∞) (x-1)(x-2)(x-3)(x-4)(x-5)/(4x-1)^a= b
分子: x^6 的系数 = 1
分母 : x^a 的系数 = 4^a
b≠0 => a=6
b= 1/4^6 =1/4096
(a,b) =(6, 1/4096)
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哥,上下指数相等
展开好麻烦😂
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