已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
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已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
![已知lim(n的平方+1/n+1+an-b)=1,求a,b的值](/uploads/image/z/3526566-6-6.jpg?t=%E5%B7%B2%E7%9F%A5lim%28n%E7%9A%84%E5%B9%B3%E6%96%B9%2B1%2Fn%2B1%2Ban-b%29%3D1%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC)
lim(n^2+1/n+1+an-b)
=lim(n^2+1/n+1+(an-b)(n+1)/n+1)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
又 lim(n^2+1/n+1+an-b)=1,可知:
[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)中[(1+a)n^2的系数1+a=0,即a=-1
故:lim(n^2+1/n+1+an-b)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
=lim[(a-b)n+(1-b)]/ (n+1)=1
则有:a-b=1,即b=a-1=-2
故:a=-1,b=-2
=lim(n^2+1/n+1+(an-b)(n+1)/n+1)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
又 lim(n^2+1/n+1+an-b)=1,可知:
[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)中[(1+a)n^2的系数1+a=0,即a=-1
故:lim(n^2+1/n+1+an-b)
=lim[(1+a)n^2+(a-b)n+(1-b)]/ (n+1)
=lim[(a-b)n+(1-b)]/ (n+1)=1
则有:a-b=1,即b=a-1=-2
故:a=-1,b=-2
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