Is this proof of, that there is no limit of lim 1/x as x approaches 0, correct? I just wanted to ask, if the following proof can be used to proof, that there is no l, such as $\lim\limits_{x \to 0} \dfrac{1}{x} = l$.
If we assume such l exists, then for every $\epsilon>0$ there is some $\delta>0$, such that for all $x$, if $\left|x \right|< \delta$, then  $\left| \dfrac{1}{x}-l\right|<\epsilon$.
Then $\left|\dfrac{1-xl}{x} \right|<\epsilon$, so  $\left|\dfrac{xl-1}{x} \right|<\epsilon$, then $\dfrac{\left|xl-1 \right|}{\left|x \right|}<\epsilon$, and finally $\left| xl-1 \right|<\epsilon\cdot\left| x \right|$.
For contradiction, we need to find $\left|x\cdot l -1 \right| \geq \epsilon\cdot \left|x \right| $.
If we choose $\epsilon =1$, then we have, for all $\delta$ that $\dfrac{1}{n}< \delta$, as $n\in N$, then we can assume $x=\dfrac{1}{n}$.
From this we have $\left|\dfrac{l}{n}-1 \right|=\left|xl-1 \right|<\epsilon\cdot \left| x \right|=\left| \dfrac{1}{n}\right|$, furthermore $\left|l-n \right|<1$, but there is certainly some combination of l,n for which $\left|x-l \right|\geq 1 = \epsilon$, which contradicts our assumption about limits.
In conclusion, there is no such l for which $\lim\limits_{x \to 0}\dfrac{1}{x}=l$ works.
 A: I don't know, but just for laughs, I'll show a proof I came up with.
Suppose there is an $l \in \mathbb{R}$ such that $\lim\limits_{x \to 0} \dfrac{1}{x} = l$.
Then there exists $\delta \in \mathbb{R}$ such that,
$\left| \dfrac{1}{x}-l\right|<0.1\ $ for all $x$ satisfying$\ 0 < \left|x \right|< \delta < \frac{1}{3}$.
Now for example, both $-\delta^2\ $ and $\ \delta^2$ are in the region from the previous line.
However, $\left| \dfrac{1}{\delta^2}-l\right|< 0.1 \implies l > 8.9$, whereas
$\left| -\dfrac{1}{\delta^2}-l\right|< 0.1 \implies l < -8.9,$ a contradiction.
A: I think you can prove this in a more straightforward way, using Cauchy's definition (proving that there's an unbounded limit - so it's not a real number) or better - you could use Heine's definition. If you choose Heine, just assume that there exists a limit, L that is a real number. Then, according to Heine, we can choose any sequence a(n) that approaches 0, as n grows to infinity, that we can substitute x with in the function ( substituting x approching to 0 with n approching infinity) and its limit will be L. Specifically, we can choose a(n) = 1/n, which will leave us with 1/(1/n)=n as n approaches to infinity - which is an unbounded sequence. This contradicts our initial assumption.
