If {$a_{n}$} is a sequence of real numbers for which $lima_{n} = a$, and if $a\neq0$, then prove that there is an $N \in \mathbb N$ so that if $n\geq N$ then $|a_{n}|\geq \frac{|a|}{2}$. Hint: Make use of the positive number $\epsilon =\frac{|a|}{2}$.
So far, I have: Since $lima_{n} = a$ and $a\neq0$, there is $N \in \mathbb N$ so that for any $n\geq N$, we have $|a_{n}-a| < \frac{|a|}{2}$. Thus for any $n\geq N$ we have $\frac{|a|}{2} > |a_{n}-a| \leq |a_{n}| - |a|$ (which we get from the Triangle Inequality).
I know that I am supposed to get $|a_{n}|$ by itself. So I can maybe say $\frac{|a|}{2} - |a| > |a_{n}|$. I am not sure how to finish my proof. Thanks in advance for any help.
You are very close. Taking the epsilon-N definition of the limit you have $$ \begin{align*} |a_n - a| < \epsilon &\iff -\epsilon < a_n - a < \epsilon \\ a - \epsilon &< a_n < \epsilon + a \\ &\text{Take $\epsilon = \frac{|a|}{2}$} \\ a - \frac{|a|}{2} &< a_n < a + \frac{|a|}{2} \\ \frac{2a - |a|}{2} &< a_n < \frac{|a| + 2a}{2} \\ \implies |a_n| &\ge \frac{|a|}{2} \end{align*} $$