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

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

Is it right?

• $L+\frac{|L|}2=\frac{L}{2}?$ That’s true if $L<0,$ but not necessarily just because $L<x.$ Nov 11, 2021 at 20:58
• @ThomasAndrews I think it's only if $x=0$ but I don't know how to continue
– Xavi
Nov 11, 2021 at 20:59
• Well.... can you do a simple change $a_n \mapsto a_n -x$? Nov 11, 2021 at 21:06
• But @Xavi, how about let another sequence, $y_n=a_n-x$? Nov 11, 2021 at 21:06
• Nov 14, 2021 at 14:53