I'm studying limits of a sequence and I'm confused as to the algebraic manipulation. My book defines an arbitrary sequence $a_n$ with $\lim a_n=s$. Then considers the definition of the limit; there exists an $N$ such that for $n>N$ $$\lvert a_n-s\rvert<\epsilon$$ then without explanation $$s-\epsilon<a_n$$ Where does this come from? How can this be said? How can the absolute value bars be dropped without knowing $a_n$ and $s$? I know intuitively its obvious but can it be shown directly?

  • 2
    $\begingroup$ Using $|a| < b \Leftrightarrow -b < a < b$. $\endgroup$ – user333870 May 28 '16 at 21:15

If $\varepsilon > 0$, $$|a_n - s| < \varepsilon \iff -\varepsilon < a_n - s < \varepsilon$$

You can attack this in cases: If $a_n - s \geq 0$, then $-\varepsilon < -(a_n-s) \leq 0 \leq a_n - s < \varepsilon$ and with appropriate modifications if $a_n - s \leq 0$.

Then, adding $s$ throughout $$ s - \varepsilon < a_n < s + \varepsilon \text{.} $$


You can do it directly: If $\epsilon > 0$ then

$$|a_n-s|<\epsilon \Leftrightarrow (a_n-s)<\epsilon, -(a_n-s)<\epsilon$$

That is,


Adding $s$ in the inequality



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.