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本帖最后由 Defluxion 于 2022-3-8 10:38 编辑
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\text{关于极限的规则}\ne \text{极限值的疑问}
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\text{方程}\phi _n\left( x \right) =x^n+x-1=0\text{在}\left( \frac{1}{2},1 \right) \text{上确定数列}\left\{ x_n \right\} ,\text{求}\underset{n\rightarrow \infty}{\lim}x_n
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\text{解:}
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\text{由零点定理、}\phi _n\left( x \right) \text{的单调性可知,}\forall n, x_n\text{的值在}\left( \frac{1}{2},1 \right) \text{存在且唯一}
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\phi _1\left( x \right) =0\Rightarrow x_1=\frac{1}{2}
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\text{设}\phi _k\left( x \right) =0\Rightarrow \left( x_k \right) ^k+x_k-1=0 ,\left( x_k \right) ^k+1=2-x_k
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\phi _{k+1}\left( x_k \right) =\left( x_k \right) ^{k+1}+x_k-1
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\,\, =x_k\left( \left( x_k \right) ^k+1 \right) -1
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\,\, =x_k\left( 2-x_k \right) -1
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\,\, =-\left( x_k+1 \right) ^2<0\text{(前一函数的解}x_k\text{代入后一函数}\phi _{k+1}(x)\text{总是小于}0\text{)}
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\phi _n\left( 1 \right) =1^n+1-1>0
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\text{知}x_1<\,\,x_2<1,x_2<x_3<1,\cdots \cdots ,x_{n-1}<x_n<1
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\text{可知}\frac{1}{2}\leqslant x_1<x_2<x_3<\cdots <x_n<1
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\text{问题:若}x_n\rightarrow 1, \text{将}1\text{代入}\phi _n\left( 1 \right) =1^n>0\ne 0,\text{不符合原方程}
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\text{这该如何解释?如果}\lim x_n\ne 1,\text{那么}\lim x_n=?
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