Showing that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist. Forgive me if this has been asked before, but I searched and could not find an answer.
I am trying to show that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist. If the limit did exist, and was equal to $l$, then for every $\varepsilon > 0$ there would exist some $\delta > 0$ such that for all $x$,
$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| < \varepsilon$$
If the limit doesn't exist, then for every number $l$, there exists some $\varepsilon > 0$ such that for all $\delta > 0$,
$$0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| \geq \varepsilon$$
So, in my efforts to find some unattainable $\varepsilon$, I have found that
\begin{align}
\varepsilon &\leq \left| \frac{1}{x} - l \right| \\\
&= \left| \frac{1}{x} - \frac{lx}{x} \right| \\\
&= \left| \frac{1-lx}{x} \right| \\\
&= \frac{|1-lx|}{|x|} \\\
&\implies \varepsilon |x| \leq |1 - lx| = |1 + (-lx)| \leq |1| + |-lx| \\\
&\implies 1 \geq \varepsilon |x| - |l| |x| = (\varepsilon - |l|) |x| \\\
&\implies |x| \leq \frac{1}{\varepsilon - |l|}
\end{align}
So I have found that if $|x| < \frac{1}{\varepsilon - |l|}$, then $\left| \frac{1}{x} - l \right| \geq \varepsilon$, for any $\varepsilon$. But, I need this to be true for any choice of $\delta$, for some specific $\varepsilon$.
Intuitively, I can think "as $\varepsilon$ gets closer to $|l|$, then the quotient becomes larger, meaning that $|x| < \frac{1}{\varepsilon - |l|}$ is satisfied for more choices of $\delta$", but I don't know how to formalize this. Any ideas for a next step? Am I completely off base with my reasoning?
 A: You're on exactly the right track. To finish up your proof, do this instead: say that for every number $L$, you'll show that for $\epsilon = 1$, you can show that for any $\delta$, there's a number $x$ with $|x - 0| < \delta$, but $|f(x) - L| > \epsilon$. 
(I'll do the remainder for the case where the putatative limit is a POSITIVE number,) 
Here's how: no matter what $\delta$ is, choose $x = \min(\delta, 1/(L+2))$. That makes $x$ positive. 
Now
$|x - 0| < \delta$ means that $x < 1/(L+2)$, so $1/x > L+2$. So $|f(x) - L|$ is at least $L+2 - L = 2 > \epsilon$. 
The small insight in this proof is that you don't need to handle every possible epsilon, and that in this case, you can reasonably easily choose one particular epsilon that will suffice for the remainder of the proof, thus cleaning things up a bit. 
(To handle the notion that the putative limit $L$ might be less than zero, you can pick $x$ as before, but use $\min(\delta, 1/(|L|+2)$. The gap between $1/x$ and $L$ will now be much larger -- like $2|L| + 2$ -- but that's fine.)
A: For the limit of a function $f$ to exist at a point $x_0$, we need the equality of the right hand limit and the left hand limit

$$ \lim_{x\to x_0^-} f(x) = \lim_{x\to x_0^+} f(x)= \lim_{x\to x_0} f(x). $$

Check also $ \liminf $ and $ \limsup $. 
A: Simpler than you think, it needs no working.
Limits must be unique from both sides (where coming in from both sides makes sense). 
It is easy to show that the limit coming in from + is different to the limit coming in from -, as limits must be unique (do you want a proof of this?) this limit does not exist.
If this is an assignment draw a picture and write "clearly the limit from the right and left differ" and move on.
A: All we need to show is that if we approach the limit from above and below you get a different answer.
$$\lim_{x \to 0^+} \frac{1}{x} = \infty
$$
As for small positive $x$ then $\frac{1}{x}$ will be positive and getting bigger as $x$ gets closer to zero.
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty
$$
As for small negative $x$ then $\frac{1}{x}$ will be negative and getting more negative as $x$ gets closer to zero.
