For some reason, I have trouble getting absolute value right. This is of a great importance in the definition of the limit.
How do I solve the following inequality for $x$:

$$|x -a| < \epsilon$$

I know the solution is $a - \epsilon < x < a + \epsilon$, but I cannot provide the logical deductions. This is how I would start:

  1. $|x-a| < \epsilon$ $\iff$ $(x-a) < \epsilon$ and $(x-a) < -\epsilon$
  2. Simplify: $(x-a) < \epsilon$ and $-(x-a) > \epsilon$

This is where I think I go wrong because now I have a negative $x$.
Could someone show me the logical deductions step by step so I can see how it turns out?


You went astray right at the start: $|x-a|<\epsilon$ is emphatically not equivalent to the statement that $x-a<\epsilon$ and $x-a<-\epsilon$. It is, as M Turgeon said, equivalent to $x-a<\epsilon$ and $x-a>-\epsilon$.

One way to see this is directly from the definition of absolute value:

$$|x-a|=\begin{cases}x-a,&\text{if }x-a\ge 0\\-(x-a),&\text{if }x-a<0\;.\end{cases}$$

If $x-a\ge 0$, then $|x-a|<\epsilon$ if and only if $x-a<\epsilon$. If $x-a<0$, then $|x-a|<\epsilon$ if and only if $-(x-a)<\epsilon$, which after multiplication by $-1$ becomes $x-a>-\epsilon$. Thus, $|a-b|<\epsilon$ if and only if $x-a<\epsilon$ and $x-a>-\epsilon$, or in one statement if and only if $-\epsilon<x-a<\epsilon$. Adding $a$ to all three ‘sides’ of this then yields the desired $a-\epsilon<x<a+\epsilon$.

Another way to see it is perhaps more intuitive. For any real numbers $a$ and $b$, $|a-b|$ is simply the distance between $a$ and $b$. The condition that $|x-a|<\epsilon$ just says that the distance between $x$ and $a$ is less than $\epsilon$. This means that $x$ must be within $\epsilon$ units on either side of $a$, i.e., between $a-\epsilon$ and $a+\epsilon$. In symbols, $a-\epsilon<x<a+\epsilon$.


By definition, the (real) absolute value function is

$$|x|=\left\{\begin{array}{}\;\;\;x&\text{ if}\,\,\,x\geq 0\\-x&\text{ if}\,\,\,x<0\end{array}\right.$$

From the above it follows that if $\,a>0\,$ ,then

$$|x|=a\Longleftrightarrow x=a\,\,\;\underline{\text{and also}}\,\,\;x=-a$$

and from here we get

$$|x|<a\Longleftrightarrow 0\leq x <a\,\,\,\;\text{and also}\;\,\,\,-a<x<0\Longleftrightarrow -a<x<a$$


This goes as follows: $$|x-a|<\epsilon \iff -\epsilon<x-a<\epsilon \iff -\epsilon+a<x<\epsilon+a,$$ where the last equivalence follows from adding $a$ everywhere.

  • $\begingroup$ I don't know the definition that you have but why do I get $(x-a) < -\epsilon$? Could you provide some steps and formal definitions? $\endgroup$ – CodeKingPlusPlus Sep 24 '12 at 3:12
  • $\begingroup$ @CodeKingPlusPlus As Brian noted above, you made a mistake when translating the definition of absolute value. My first equivalence is the definition of absolute value (or if you like the case-by-case definition used by both Brian and DonAntonio, then I used both the definition and the equivalence $-(x-a)<\epsilon\iff(x-a)>-\epsilon$). $\endgroup$ – M Turgeon Sep 24 '12 at 14:24

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