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Let us say we found a $\delta$ for some limit $f(x)$ as $x$ goes to $a = l$.

So we have $ |x - a| < \delta(\epsilon) $, but it is also true that $|x-b| <\delta $ for some $b$ less than $a$ but greater than $0$.

Now I'll claim that the limit as $x$ tends to $b$ is $l$. What's wrong with what I did right now? For every $\epsilon$ this same $\delta$ would work.

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    $\begingroup$ not sure about others, I have difficulty understanding the questions. $|x-a| < \delta (\epsilon) 0$? $\endgroup$ – Siong Thye Goh Jul 19 '17 at 18:22
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    $\begingroup$ It is not true that if $|x-a| < \delta$ and $b < \delta$ then $|x-b| < \delta$, as you seem to be implying in your third line. $\endgroup$ – астон вілла олоф мэллбэрг Jul 19 '17 at 18:23
  • $\begingroup$ @siong, Ay that 0 was a mistake $\endgroup$ – Vrisk Jul 19 '17 at 18:30
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    $\begingroup$ Take $a=0$, $b=-100$, $\delta=1$, and $x=1/2$. We have $|x-a| < \delta$, but $|x-b| \ge \delta$, even though $b<a$. $\endgroup$ – angryavian Jul 19 '17 at 18:34
  • $\begingroup$ @angryavian, make b greater than 0 also $\endgroup$ – Vrisk Jul 19 '17 at 18:35
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I found your question somewhat difficult to understand, so please let me know if my response helps. The reason this doesn't work is basically entirely encapsulated by what $\lvert x-x_0\rvert$ really is. $\lvert x-x_0\rvert$ denotes the distance from $x$ to $x_0$ on the real line. When we discuss continuity of a function $f:\mathbf{R}\to\mathbf{R}$ at a point $a\in \mathbf{R},$ we are interested in understanding what happens to $f(x)$ as $x$ gets very close to $a$. Unless we are in a special case where the function $f$ behaves in some sort of global pattern, the behavior of $f$ near $a$ will not tell us anything about the behavior of $f$ near $b$, if $a\ne b$.

The idea with continuity is that given $\epsilon>0$, we can find a $\delta$ quite small so that $\lvert x-a\rvert<\delta$ implies that $\lvert f(x)-L\rvert<\epsilon$, where $L=\lim_{x\to a}f(x)$.

However, different pairs $(a,\epsilon)$ and $(b,\epsilon)$ may demand different $\delta$'s. For instance, given the function $f(x)=x^2$, let $a=0, b=1, \epsilon=1$. Now, we know that $f$ is continuous, so in particular it is continuous at $a$ and $b$. At $a=0$, if we want $\lvert f(x)-f(0)\rvert=\lvert f(x)\rvert<1$ we can choose $\delta=1$ and stipulate that $\lvert x-0\rvert<\delta=1$. On the other hand, at $b=1$, $\delta=1$ will not work, because $\lvert x-1\rvert<1$ does not imply that $\lvert f(x)-f(1)\rvert<1$ since if $x=1.5$, $\lvert 1.5-1\rvert<1$ but $\lvert f(1.5)-f(1)\rvert=\lvert 2.25-1\rvert=1.25>1$. As such, we can see that we need to choose a smaller $\delta$. $\delta=.1$ will work, for instance.

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  • $\begingroup$ Ahh the old a implies b but b doesn't imply a. I see what happened. The things within the distance but that distance doesn't mean anything. Thank you $\endgroup$ – Vrisk Jul 19 '17 at 19:09
  • $\begingroup$ Hey I'm confused again. When you prove a limit, you prove |x-a|< delta. Can't you do |x|-|a| < delta and add a to delta, thus 'proving' limit near 0? $\endgroup$ – Vrisk Jul 20 '17 at 5:56
  • $\begingroup$ What limit are you taking? $\endgroup$ – Antonios-Alexandros Robotis Jul 20 '17 at 15:28
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This statement

So we have $ |x - a| < \delta(\epsilon) $, but it is also true that $|x-b| <\delta $ for some $b$ less than $a$ but greater than $0$.

is false for some fixed $x$. Just take, by example, $a=1$, $b=1/2$, $x=2$ and $\delta=1.1$.

If we assume that $x$ is not fixed then the (order) relation between the sets $$A:=\{x\in\Bbb R:|x-a|<\delta\},\quad B:=\{x\in\Bbb R:|x-b|<\delta\}$$

can be arbitrary, that is for suitable $a$ and $b$ is possible that $A\subset B$ or $B\subset A$ or $A\cap B=\emptyset$, etc...

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  • $\begingroup$ But if |a-x|<\delta then doesn't |x|-|a|<\delta, always? $\endgroup$ – Vrisk Jul 20 '17 at 5:54
  • $\begingroup$ @Vrisk yes, this is correct. $\endgroup$ – Masacroso Jul 20 '17 at 5:55
  • $\begingroup$ So |x-0| < delta + a. Therefore limit near 0 is L???? $\endgroup$ – Vrisk Jul 20 '17 at 6:00
  • $\begingroup$ @Vrisk Yes, we have that $|x-a|<\delta\implies |x|<\delta+a$ for $a>0$ but I dont follow exactly your complaint or how it is related to my answer. What limit are you talking about, limit of what function at what point? $\endgroup$ – Masacroso Jul 20 '17 at 7:03
  • $\begingroup$ just for the record to clear things, we have that $$\lim_{x\to a}f(x)=L\iff\forall\epsilon>0\exists\delta>0\forall x\in\operatorname{dom}(f):0<|x-a|<\delta\implies |f(x)-L|<\epsilon$$ when $|L|<\infty$. $\endgroup$ – Masacroso Jul 20 '17 at 7:19
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"So we have |x−a|<δ(ϵ), but it is also true that |x−b|<δ for some b less than a but greater than 0."

But that isn't true.

Just take any $x$ so that $b + \delta < x < a+\delta$. Then $|x-b| > \delta$.

But $|x - a|$ can very easily be less than $\delta$.

Let $\max(b+ \delta, a - \delta) < x < a+\delta$ and we have both $|x -b| > \delta$ and $|x-a| < \delta$.

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(Actually if $b + \delta < a-\delta$ than $|x -a| < \delta \implies |x-b| > \delta$ always, so we didn't need to speculate $\max(b+\delta,a-\delta)< x< a+\delta$. Simply stating $b + \delta < x < a+\delta$ would have been enough.)

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$|x-a|<\delta \iff a-\delta <x<a+\delta$ and for the limit that means this inequality $|f(x)-\ell|<\varepsilon$ is satisfied $\forall x \in(a-\delta,a+\delta)$ or the suffiscent condition to satisfy $|f(x)-\ell|<\varepsilon$ is $x \in(a-\delta,a+\delta)$

$|x-b|<\delta \iff b-\delta <x<b+\delta$ and for $x \in(b-\delta,b+\delta)\implies \left\lbrace\begin{array}l |f(x)-\ell|<\varepsilon\\\text{or} \\ |f(x)-\ell|\geq\varepsilon \end{array} \right.$

A reminder below : enter image description here

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