Prove that $|x|<\epsilon$ for every $\epsilon$ greater than $0$ if and only if $x$ is equal to zero. Prove that $|x|<\epsilon$ for every $\epsilon$ greater than $0$ if and only if $x$ is equal to zero.
My Attempt:
Assume $x$ to be a non-zero number, say $x=2$. Clearly there is a contradiction here if $\epsilon=1$
If $x=0$ then $\epsilon$ can be any positive number.
Is my reasoning correct or some more detail is required.
I have just begun to study $\epsilon-\delta$ definition of limit and was given this problem to start with.
 A: If $x\neq 0$, then $|x|> 0$. Take
$$ 0<\varepsilon = \frac{|x|}{2}< |x|.$$
A: The if part
$|x|<\epsilon \implies -\epsilon<x<\epsilon$
$\therefore 0<x<\epsilon$  ,now suppose $x>0$ and let $\epsilon_{1}=\frac{x}{2}>0$
then $0<\epsilon_{1}<x$ which is false, hence $x=0$
A: If you have saw the sandwitch thm, say that if $g \leq f \leq h$ and $\lim_{x\to x_0} g(x)=\lim_{x\to x_0} h(x)$ then $\lim_{x\to x_0} f(x)=\lim_{x\to x_0} h(x)$, you can use the fact that $0 \leq|x|\leq \frac{1}{n}$ , which holds for every natural $n$, and when $n$ goes to infinity , both limits are $0$.
A: 
Prove that $\;|x|<\varepsilon\;$ for every $\,\varepsilon\,$ greater than $\,0\,$ if and only if $\,x\,$ is equal to zero.

First, we will prove that
if $\;|x|<\varepsilon\;$ for all $\,\varepsilon>0\,,\,$ then $\,x=0\,.$
Indeed, if $\,x\,$ were not equal to zero, then there would exist $\,\varepsilon=|x|>0\,$ such that $\,|x|\not<\varepsilon\,,\,$ but it is a contradiction. Hence, $\,x\,$ has to be equal to zero.
Now, we will prove that
if $\,x=0\,,\,$ then $\;|x|<\varepsilon\;$ for all $\,\varepsilon>0\,.$
Indeed, $\,|x|=0<\varepsilon\;$ for any $\,\varepsilon>0\,.$
