clarification on showing $\lim_{x\to 0} \frac{1}{x}$ does not exist. So this has to do with the classic problem from here: Proof that the limit of $\frac{1}{x}$ as $x$ approaches $0$ does not exist
I only had a question of how to properly come up with the needed contradiction. So if I am proving this via contradiction I would be assuming that $\left|\frac{1}{x} - l\right| < \epsilon$,  my issue is, how do I arrive at the necessary $x$ because to obtain that $$\left|\frac{1}{x}\right| > \epsilon + |l|$$ I would need to use the condition $$\left|\frac{1}{x} - l\right| > \epsilon$$  
Doing the manipulations with my current assumption is not leading me to the right answer. What would I have to do to arrive at the contradiction?
 A: First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x$ if $0 < \lvert x\rvert < \delta$ then $\lvert{1\over x} - L \rvert <\epsilon$. Now, let $\epsilon=1$ so that $0 < \lvert x \rvert < \delta$ implies $\lvert{1\over x} - L \rvert < 1$. In order to get a contradiction, we need $\lvert{1\over x}-L\rvert \geq 1$. In order to do this, we now split the proof up into two cases.
Case 1: $L \geq 0$
Since $L$ is non-negative, if we get $\frac 1 x \geq L+1$, we will have $\lvert\frac 1 x-L \rvert \geq 1$, which gives us our contradiction. We need $0 < x < \delta$ in order to satisfy the hypothesis and $0 < x \leq \frac{1}{L+1}$ in order to get the contradiction. In order to get $x$ in both of these intervals, we will choose $x=\min(\frac{\delta}{2}, \frac{1}{L+1})$. Now that we have $0 < x \leq \frac{1}{L+1}$, we can deduce $\frac 1 x \geq L+1$, or that $\frac 1 x-L \geq 1$, or that $\lvert\frac 1 x-L \rvert \geq 1$, which gives us our contradiction.
Case 2: $L < 0$
Since $L$ is negative, if we get $\frac 1 x \leq L-1$, we will have $\lvert\frac 1 x-L \rvert \geq 1$, which gives us our contradiction. We need $0 > x > -\delta$ in order to satisfy the hypothesis and $0 > x \geq \frac{1}{L-1}$ in order to get the contradiction. In order to get $x$ in both of these intervals, we will choose $x=\max(-\frac{\delta}{2}, \frac{1}{L-1})$. Now that we have $0 > x \geq \frac{1}{L-1}$, we can deduce $\frac 1 x \leq L-1$, or that $\frac 1 x-L \leq -1$, or that $\lvert\frac 1 x-L \rvert \geq 1$, which gives us our contradiction.
A: Suppose the limit $L$ exists and is finite. Then, for every $\varepsilon>0$, there is $\delta>0$ such that, for $0<|x|<\delta$,
$$
\left|\frac{1}{x}-L\right|<\varepsilon
$$
Then, for $0<|x|<\delta$, 
$$
-\varepsilon+L<\frac{1}{x}<\varepsilon+L
$$
Now observe that
$$
-\varepsilon-|L|\le-\varepsilon-L
\quad\text{and}\quad
\varepsilon+L\le\varepsilon+|L|
$$
so
$$
-\varepsilon-|L|<\frac{1}{x}<\varepsilon+|L|
$$
and so
$$
\left|\frac{1}{x}\right|<\varepsilon+|L|
$$
implying
$$
|x|>\frac{1}{\varepsilon+|L|}
$$
in contradiction with $0<|x|<\delta$.
A: Using the Sequential Criterion: 
If $\lim_{x\to 0} \frac{1}{x}=L$ exist, so every sequence $x_n \in \mathbb{R}- \{ 0\}$ where $\lim_{n-\infty}  x_n = 0$ then $\lim_{n\to \infty} f(x_n)=L$.
Get:
$x_n=\frac{1}{n}$ clearly $x_n \rightarrow 0$ but $f(x_n)=n \rightarrow \infty$ so $lim_{x\to 0} \frac{1}{x}$ does not exist
