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I have a problem similar to the one answers in this post (Mhenni Benghorbal's answer)

$f(x)=x^2$, find $\delta$ such that $|x-1|\leq \delta$ implies $|f(x)-1|\leq \epsilon$.

I am copying part of his solution:

We must show that for every $\epsilon >0$ there is $\delta >0$ such that if $0<|x-1|<\delta\,,$ then $|x^2-1|<\epsilon$. Finding $\delta$ is most easily accomplished by working backward. Manipulate the second inequality until it contains a term of the form $x-1$ as in the first inequality. This is easy here. First $$ |x^2-1|=|x+1||x-1| \,. $$ In the above, there is unwanted factor of $|x+1|$, that must be bounded. If we make certain that $\delta<1$ $$ |x-1|<\delta<1 \,,$$ then $$ |x-1|< \delta \implies |x-1|< 1 \implies -1<x-1<1 \,$$ Adding $2$ to the last inequality gives $$ 1<x+1<3 \implies |x+1|<3\,.$$ So, if $$ |x^2-1|=|x+1||x-1|<3|x-1|<\epsilon \implies |x-1|<\frac{\epsilon}{3}\,. $$ Now, select $\delta = \mathrm{min}\left\{ 1, \frac{\epsilon}{3}\right\} $.

Please clarify: why do we set $\delta=1$ and not another integer? Why must we take the minimum? The $\epsilon/3$ part makes sense, but I completely don't understand how the 1 fits into this solution. Please help!

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    $\begingroup$ No, you got that wrong, $\delta < 1$ it's not set to 1. It's set to 1 if $\frac{\epsilon}{3}\ge 1$, but if it's smaller then it's set to $\frac{\epsilon}{3}$. $\endgroup$ – mjb4 Feb 2 '14 at 17:08
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Let's go the other way. The answer is: for every $\varepsilon>0$ there is $\delta=min\left\{1,\frac{\varepsilon}{3}\right \}$ such that if $0<|x-1|<\delta$, then $|x^2-1|<\varepsilon$.

You can go this route and verify that it is true?

If you can not do this read here:

Let $\varepsilon>0$.

First place verify that $\delta=\varepsilon/3$ or $\delta=\min\left\{1,\varepsilon/3\right\}$.

(i) Suppose that $\varepsilon/3>1$, so $\delta=1$ $$ |x-1|<\delta=1 \Rightarrow |x-1|<\delta$$ then $$ |x-1|<1 \Rightarrow -1<x-1<1 $$

Adding 2 to the last inequality gives

$$ 1<x+1<3 \Rightarrow |x+1|<3 $$

So,

$$ |x^2-1|=|x-1||x+1|<3|x-1| $$

Here, remember that $\varepsilon/3>1$, so as $|x-1|<1$, $|x-1|<1<\varepsilon/3$. What I have now

$$|x^2-1|=|x-1||x+1|<3|x-1|<3(\varepsilon/3)=\varepsilon$$

(ii) If $\varepsilon/3<1$, we have the case that you are convinced

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  • $\begingroup$ Thank you, this makes sense, but how does one know to come up with 1? I saw another example with $|x^2-4|$ and there we also set $\delta<1$. Do we always include 1 in the minimum? $\endgroup$ – user989 Feb 2 '14 at 17:41
  • $\begingroup$ @user989 Not always! In the case of this exercise worked because assuming that $ | x-1 | <$ 1 "things worked out." It is a type of tactics you'll have to get used to. Look what value makes things work to get where you want. Do not hold on to it, one thing you have to remember is that mathematicians are lazy, so always work with simple values​​: $-1, 1, 0 ...$ :P $\endgroup$ – Felipe Feb 2 '14 at 17:57
  • $\begingroup$ Thank you! Just to clarify: if $\epsilon/3 <1$, is it the same solution, except we write $|x-1|<\epsilon/3<1$ and proceed as before? I thought I understood the case, but now I'm getting a bit confused about the need for 2 values in the min. $\endgroup$ – user989 Feb 2 '14 at 18:02
  • $\begingroup$ @user989 Yes, you are right! Think of it this way: the function $\min\{a,b\}$ is a function that answers the question about the set in there " which of those elements is the smallest?". In our case, it follows that $\min\{1,\varepsilon/3\}=\varepsilon/3$ if $\varepsilon/3<1$ $\endgroup$ – Felipe Feb 2 '14 at 18:08
  • $\begingroup$ Thank you very much for your time! I think the source of my confusion is this: if in both cases we end up ultimately plugging in $\epsilon/3$, why don't we just take $\delta=\epsilon/3$, and not worry about whether it's greater or less than 1, 2 or another number? $\endgroup$ – user989 Feb 2 '14 at 18:18

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