Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? 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:
$$0<|x|<\delta \implies |x|>\frac{|1-xl|}{\varepsilon}$$ 
 A: 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 $x=\frac12\min\{\frac1{|l|+\epsilon},\delta\}$.
A: 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 $x=0$.
A: 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 :-)
