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I am seriously struggling with $\varepsilon - \delta$ proofs overall continuity of a function (e.g. here there is an attempt).

In the attempt to see how to prove the continuity of $f(x) = x^2 - 2$, I got to the point where I assume $\delta \leq 1$, and I proceed as follows:

\begin{align*} | x - c | < 1 & \Longleftrightarrow -1 < x - c < 1 \\ & \Longleftrightarrow c-1 < x < c+1. \end{align*}

If we add $c$ to the previous expression we obtain

$$ 2c -1 < x + c < 2c + 1 \Longleftrightarrow | x + c | < 2c+1.$$

But this does not work, because I should get $2|c|+1$.

What am I missing?

I think I truly miss a basic understanding of how absolute value works in this context (plus some other things, but in this question I would focus on the absolute value issue).

Thanks a lot for your time as always.

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    $\begingroup$ Does it help to note that $2c+1<2|c|+1$? $\endgroup$ – Servaes Jun 6 '16 at 18:12
  • $\begingroup$ Thanks, it does a lot! (Shame on me) $\endgroup$ – Kolmin Jun 6 '16 at 18:17
  • $\begingroup$ Decompose the last absolute value and see that the equality that you put isnt true. $\endgroup$ – Masacroso Jun 6 '16 at 18:18
  • $\begingroup$ *Should be a $\leq$, shame on me ;) $\endgroup$ – Servaes Jun 6 '16 at 18:19
  • $\begingroup$ @Masacroso: Are you referring to mine inequality, or to Servaes' one? $\endgroup$ – Kolmin Jun 6 '16 at 19:09
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Note that if $c<0$, $c=-|c|$. Then, for $c<0$, we have

$$-(2|c|+1)<x+c<(2|c|+1)$$

Hence, $|x+c|<|\,(2|c|+1)\,|$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Aug 12 '16 at 1:57

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