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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.

Maverick
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    What happens if $x=1/2$? Now your choice of $\epsilon$ must be different to reach a contradiction. The way to fix the proof is to define exactly your $\epsilon$ for all choices of $x \neq 0$. – AnilCh Jun 28 '21 at 11:19
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    The comment of @AnilCh raises a very good point. Focus on your question's first sentence. The issue is, for which $x$ is it true that no matter what $\epsilon > 0$ is chosen, $|x| < \epsilon.$ If $x = 1$, then you can choose $\epsilon = (1/2).$ However, if $x = (1/10)$, clearly, the choice of $\epsilon = (1/2)$ doesn't work. Therefore, you must specify the choice of $\epsilon$ in terms of $x$. – user2661923 Jun 28 '21 at 11:24
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    Can you observe that your problem is equivalent to "$0$ is smallest non negative real number"? Also the quoted statement remains true if "real" is replaced by "rational". You should compare it with "$1$ is the smallest positive integer". – Paramanand Singh Jun 28 '21 at 13:50
  • Does this answer your question? Intuition about : $a = b \iff | a − b| < \epsilon$, for every $\epsilon > 0$ –  Sep 22 '23 at 08:58

4 Answers4

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

amnesiac
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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 $x=0$

Shubham
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    If $,0<x<\frac{\sqrt2}2,,,$ then $,\varepsilon_1>x;.$ Why did you write that $,0<\varepsilon_1<x;?$ – Angelo Jun 28 '21 at 12:43
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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 $0$.

Angelo
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Ron Abramovich
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  • Clearly this has nothing to do with limits or Sandwich theorem. Here we are talking about a specific property satisfied by a given real number $x$ and we are supposed to prove that this holds only when $x=0$. Moreover the idea is related to order relations and holds in contexts of rational numbers too. – Paramanand Singh Jun 28 '21 at 11:47
  • I don't agree. I think that , as the question arises, it comes from a guy in the early math career. That way of thinking about that situation gives intuitive explenation for that idea. One can think of that given number as a constant number in order to adjust it to the thm. – Ron Abramovich Jun 28 '21 at 11:50
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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 $\,\varepsilon>0\,.$

Angelo
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