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Suppose that $\lim_{n\rightarrow \infty} a_n = L$ and $L \neq x$. Prove there is some $N$ such that $a_n \neq x$ for all $n > N$.

We know by the definition of convergence of a sequence, $\forall \epsilon > 0, \exists\ N \in \mathbb{N}$ such that $\forall n \geq N$, $|x_n - L| < \epsilon$.

I try to chose $\epsilon = \frac{\left|L\right|}{2}$ and then:

  1. If $L<x$ we have $L+\frac{|L|}2=\frac{L}2<x.$ Hence for all $n\ge N$, $a_n<x$.
  2. If $L>x$ we have $L-\frac{|L|}2=\frac{L}2>x.$ Hence for all $n\ge N$, $a_n >x$.

Is it right?

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