Using $\varepsilon - \delta$, how would I show $\lim\limits_{x \to 0}f(x)$ has no limit? 
$\text{Does }\lim\limits_{x \to 0}\frac{7}{x}\text{ exist?}$

I know that the limit doesn't exist, but don’t know how to prove it. I don’t really understand how to use the $\varepsilon -\delta$ definition of a limit to solve this. Any pointers would be helpful.
 A: For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that
$0<|x|<\delta\implies\big|f(x)\big|=\left|\dfrac{7}{x}\right|>\dfrac{7}{\delta}=7\dfrac{\epsilon}{7}=\epsilon\;.$
Hence there exists $\;\lim\limits_{x\to0} \big|f(x)\big|=\lim\limits_{x\to0} \left|\dfrac{7}{x}\right|=+\infty\;.$
Analogously we can prove that there exist
$\lim\limits_{x\to0^+} f(x)=\lim\limits_{x\to0^+}\dfrac{7}{x}=+\infty\;,\quad\color{blue}{(*)}$
$\lim\limits_{x\to0^-} f(x)=\lim\limits_{x\to0^-}\dfrac{7}{x}=-\infty\;.\quad\color{blue}{(**)}$
Now we are going to prove the limit $\;(*)\;$.
For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that
$0<x<\delta\implies f(x)=\dfrac{7}{x}>\dfrac{7}{\delta}=7\dfrac{\epsilon}{7}=\epsilon\;.$
Now we are going to prove the limit $\;(**)\;$.
For any $\;\epsilon>0\;$ there exists $\;\delta=\dfrac{7}{\epsilon}>0\;$ such that
$-\delta<x<0\implies f(x)=\dfrac{7}{x}<-\dfrac{7}{\delta}=-7\dfrac{\epsilon}{7}=-\epsilon\;.$
We have just proved the limits $\;(*)\;$ and $\;(**)\;.$
Since the results of the limits $\;(*)\;$ and $\;(**)\;$ are different $(+\infty$ and $-\infty)$, there does not exist the limit $\;\lim\limits_{x\to0}f(x)=\lim\limits_{x\to0}\dfrac{7}{x}\;.$
A: The idea is this: you want to show that as $x\to0$, you can make $|f(x)|$ arbitrarily large, which is to say, bigger than a given value $\epsilon$. Think of it as a game. I choose, say, $\epsilon = 70$. Can you find a neighbourhood of $0$ for which $|f(x)|$ is guaranteed to be greater than $70$ when $x$ is in that neighbourhood? In other words, how small would $\delta$ have to be, where the neighbourhood is $0 < |x| < \delta$?
If you can do that, then what if I choose $\epsilon = 7000000$? How small would $\delta$ have to be in that case?
What if I choose an arbitrary $\epsilon > 0$?
A: If this limit exists, let us call him $L$. By definition, $\forall\varepsilon>0$ there exist $\delta>0$ such that, $x\neq0$ and $|x-0|=|x|<\delta\implies|\frac{7}{x}-L|<\varepsilon$. Taking, for example, $\varepsilon=1$, there exist $\delta=\delta(1)$ such that $$x\neq0, |x|<\delta\implies|\frac{7}{x}-L|<1(*).$$ On the other hand, Let $n$ be the least positive integer $m\in\mathbb{N}$ such that $\frac{1}{m}<\delta$. Observe that we obtain the increasing sequence $$f(\frac{1}{n})=7n<f(\frac{1}{n+1})=7(n+1)<f(\frac{1}{n+2})=7(n+2)<\cdots$$
which clearly will overcome $L+1$. But, for all $m\geqslant n$, we have $\frac{1}{m}\leqslant\frac{1}{n}<\delta$, and we get a contradiction for $(*)$.
