Prove that $|x|<\epsilon$ for every $\epsilon$ greater than $0$ if and only if $x$ is equal to zero.

Question

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.

Answer 1

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 $\,\varepsilon>0\,.$

Answer 2

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 $0$.

Answer 3

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 $x=0$

Answer 4

If $x\neq 0$, then $|x|> 0$. Take $$ 0<\varepsilon = \frac{|x|}{2}< |x|.$$