How to prove that limit doesn't exist using epsilon-delta definition?

It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.

Please show through an example (you may show that $\lim_{x\rightarrow0} \frac{1}{x}$ doesn't exist)

If possible please use the explanation scheme that is used by this answer https://math.stackexchange.com/a/66552/335742

Suppose $L=\lim_{x\to0}\frac{1}{x}$ is finite.

If $L>0$, then there exists $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}-L|<L$, that is, $$ 0<\frac{1}{x}<2L $$ This is a contradiction, just take $-\delta<x<0$.

Similarly if $L<0$.

Thus we can only have $L=0$. Then there should exist $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}|<1$, an obvious contradiction.

It can be neither $\lim_{x\to0}\frac{1}{x}=\infty$ nor $\lim_{x\to0}\frac{1}{x}=-\infty$, because $\frac{1}{x}$ assumes positive and negative values in every punctured neighborhood of $0$.

Can this be generalized? Not really. For instance, in order to show the non existence of $\lim_{x\to0}\sin\frac{1}{x}$ the easiest way is to show that the limit should be in the interval $[-1,1]$, but that $\sin\frac{1}{x}$ assumes every value in $[-1,1]$ in each punctured neighborhood of $0$, so it is far from every possible limit. Alternatively, there exist sequences $(a_n)$ and $(b_n)$ convergin to $0$ such that $\lim_{n\to\infty}\sin\frac{1}{a_n}=0$ and $\lim_{n\to\infty}\sin\frac{1}{b_n}=1$.

As another example, $\lim_{x\to0}e^{1/x}$ doesn't exist because the one sided limits are different: from the left it is $0$, from the right it is $\infty$.

If $\displaystyle\lim_{x\to0} \frac 1 x = L$ then what happens if $\varepsilon=1$? There would exist $\delta>0$ such that if $0<|x|<\delta$ then $\displaystyle \left| \frac 1 x - L \right| <\varepsilon.$ But if $x$ is positive, you can make $1/x$ bigger than $L+\varepsilon$ by making $x$ less than $1/(L+\varepsilon).$ And a similar thing works if $x$ is negative (where you have $1/x\to-\infty$).