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The $\varepsilon-\delta$ definition of limit is:

$$\lim_{x \to a}f(x)=l \iff (\forall \varepsilon>0)(\exists \delta>0)(\forall x \in A)(0<|x-a|<\delta \implies |f(x)-l|<\varepsilon)$$

Consider there is $\lim\limits_{x \to 0}f(x)=l$ for $f(x)=1/x$. Then:

$$(\forall \varepsilon>0)(\exists \delta>0)(\forall x \in R-\{0\})(0<|x|<\delta \implies |1/x -l|<\varepsilon)$$

Can we prove that there is no limit just sticking to the above statement without using any other theorem? The manipulation gives the following but I can't figure it out:

$$0<|x|<\delta \implies |x|>\frac{|1-xl|}{\varepsilon}$$

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  • $\begingroup$ Are you really sure there is no limit? $\endgroup$ – Caligula Feb 14 '15 at 11:34
  • $\begingroup$ @Alessandro If there is we can be able to prove it too. $\endgroup$ – user137035 Feb 14 '15 at 11:42
  • $\begingroup$ Indeed the limit exists, but you are not using the correct definition (or, if you prefer, you're looking for the wrong kind of limit) $\endgroup$ – Caligula Feb 14 '15 at 11:46
  • $\begingroup$ @Alessandro, what is that limit? $\endgroup$ – Arashium Feb 14 '15 at 12:06
  • $\begingroup$ Alessandro: By definition (the one that taught me) a limit exist only if (a)the function is continius at that point, (b)the value is some finite number $l \in \Bbb{R}$ and (c) both lateral limits coincide and are equal to $l$. According to this definition, $f(x)=1/x$ does not have a limit at $x=0$. Am I missing something? Please, correct me if I'm wrong. $\endgroup$ – Карпатський Feb 14 '15 at 12:14
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Assume $l$ is a limit and for some $\epsilon>0$ you have found some $\delta>0$. Then $|f(x)-l|<\epsilon$ whenever $0<|x|<\delta$ implies that $f$ is bounded on $(-\delta,\delta)\setminus\{0\}$, whihc is not the case.

Alternatively and explicitly, no matter what $\delta$ you pick for given $\epsilon>0$, we have $|\frac1x-l|>\epsilon$ at least for $x=\frac12\min\{\frac1{|l|+\epsilon},\delta\}$.

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Let $M>0$ and $N=\frac{1}{M}$. If $0<x<M$ you have that $f(x)>M$ and thus $f(x)\to +\infty $ if $x\to 0^+$. If $-M<x<0$ you have that $f(x)<-M$ and thus $f(x)\to-\infty $ if $x\to 0^-$, therefore there is no limit on $x=0$.

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  • $\begingroup$ You are using one sided limit definitions. $\endgroup$ – user137035 Feb 14 '15 at 13:21
  • $\begingroup$ Yes I do, but in the both side (it mean in $0^-$ and $0^+$). And then we can conclude. $\endgroup$ – Surb Feb 14 '15 at 13:29
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The task is for any given number $l$ and for any given tolerance challenge $\epsilon$ to find a positive $\delta$ that will for all $x \in (-\delta,0) \cup (0, \delta)$ keep $$ \left\lvert \frac{1}{x} - l \right\rvert < \epsilon \quad (*) $$ The problematic item is the $0$ end of the intervals where to choose the $x$ from, independent of $\delta$. This allows for arbitrary small $\lvert x \lvert$, which means arbitrary large $\lvert 1/x \rvert$, which can surpass any given number $l$, so the difference in $(*)$ can not be bound by any number $\epsilon$.

(Hagen used less words for the same phenomenon :-)

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  • $\begingroup$ Your answer really helped me to understand what Hagen is telling but I had to choose his answer because he has an alternative explanation. Thank you very much. $\endgroup$ – user137035 Feb 14 '15 at 13:31
  • $\begingroup$ No problem with that. $\endgroup$ – mvw Feb 14 '15 at 13:33

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