If {$x_{n}$} is a sequence of real numbers for which $\lim x_{n} = x$, and if $x\neq0$, then prove that there is an $N \in \mathbb N$ so that if $n\geq N$ then $|x_{n}|\geq \frac{|x|}{2}$. Hint: Make use of the positive number $\epsilon =\frac{|x|}{2}$. You may need to use the triangle inequality for a difference.
So far, I have: Let $\epsilon > 0$ be arbitrary. Since $\lim x_{n} = x$, applying the definition of a limit with $\epsilon = \frac{|x|}{2}$, we have an $N \in \mathbb{N}$ such that $|x_{n}-x|< \frac{|x|}{2}$ for any $n \geq N$.
So I know that as long as I can prove that $|x_{n}|> \frac{|x|}{2}$ then it sufficient. However. I am not sure where to use the given hint of the triangle inequality. Any help is appreciated. Also please explain each step after what I have.
By definition of limit, we have
$|x_n - x| < \epsilon$ for $n \ge N$
But $|x| - |x_n| \le ||x| - |x_n|| \le |x_n - x|$ by reverse triangle inequality
Putting the two together,
$|x_n| > |x| - \epsilon$
now set epsilon as mentioned above since this holds for all positive epsilon