Proof for $| x_n | >$ $ 1 \over 2$ $|a|$

Question

Suppose $x_n → a$ when $n → ∞$. Suppose further that $a ≠ 0$. Prove that there exists $K ∈ N$ for which $\forall$ $n > K$ holds:

$| x_n | >$ $ 1 \over 2$ $|a|$

This is one of those kinds of epsilon proofs that seems a bit wierd to me. It seems that you cannot just start looking at the distance $| x_n - a|$ and find the "equivalence" with $\epsilon$ here like you usually do. I see that to prove this we would need to use the triangle inequality, but how does one approach this kind of problem?

Answer 1

There exist $K$ such that $|x_n-a|<\frac {|a|} 2$ whenever $n >K$. Now use the inequality $|a| \leq |a-x_n|+|x_n|=|x_n-a|+|x_n|<\frac {|a|} 2+|x_n|$ to get $|x_n|>\frac {|a|} 2$ for $n >K$

Answer 2

Since $\lim_{n\to\infty}x_n=a$ and since $\frac{\lvert a\rvert}2>0$, there is some $N\in\mathbb N$ such that$$n\geqslant N\implies\lvert x_n-a\rvert<\frac{\lvert a\rvert}2.$$But then, if $n\geqslant N$,$$\lvert x_n\rvert=\lvert x_n-a+a\rvert\geqslant\lvert a\rvert-\lvert x_n-a\rvert>\lvert a\rvert-\frac{\lvert a\rvert}2=\frac{\lvert a\rvert}2.$$