I wanted to disprove the existence of limit;
$$\lim_{x\to 0} {1\over x} $$
I proved it in the following way.
Let us suppose to the contrary that the limit exists.
We consider 2 cases.
(i) $L≠0$
Let $ε=L/4.$
$0<|x|<δ ⇒ |(1/x)-L| < L/4$,
Let $x=1/(2L)$ such that $$0<|1/2L|<δ
⇒ |2L-L|<L/4 ⇒ |L|<L/4,$$
which is a contradiction.
(ii) $L=0.$
Let $ε=1.$
$0<|x|<δ ⇒ |(1/x)| < 1$,
Let $x=1/2$ such that $$0<|1/2|<δ
⇒ |2|<1,$$
which is a contradiction.
Hence the existance of limit is disproved.
I am new to advanced mathematics and mathematical proofs. I don't know whether this proof is correct or not. It would be helpful if someone could verify whether this is correct and, if not, could point out the mistakes. Thank you.
answer for case (i)
Choose $\epsilon = \frac{|L|}{4}.$ I do not think you are free to assume $\frac1{|2L|}<\delta$, but we can do something similar. We may choose $x>0$ such that both $0<x<\delta$ and $0<x<\frac1{|2L|}$ hold, or written together, $$0<x<\min\left(\delta,\frac1{|2L|}\right).$$
From our epsilon-delta implication, we may now write $$\left|\frac1x\right| - \left|L\right| \leq \left|\frac1x - L\right|<\frac{|L|}4,$$
$$0<\frac1x \leq \frac{5|L|}4,$$
$$0< \frac4{5|L|} \leq x < \frac1{2|L|}.$$ This last relation is not possible.
hint for case (ii)
Your proof for the second case is flawed in the same way as your proof for the first case, but you can fix it in the same way: by choosing your positive $x$ to satisfy two inequalities simultaneously.