I would like to understand how to complete this proof, for I am unable to conclude. Also I wanted to know if what I wrote before is correct. I tried to follow an answer over here (this one: How to prove a limit exists using the $\epsilon$-$\delta$ definition of a limit and the answer I followed is by the user FREETRADER).
So I want to prove $\lim_{x\to 2} 3x = 7$ is false.
Following the linked example, let $\epsilon > 0$.We have to demonstrate that there does exists $\delta > 0$ such that IF $0 < |x-2| < \delta$ THEN $|f(x) - \ell| < \epsilon$.
In what way the size of $|x-2|$ afflicts the size of $|f(x) - 7|$? Since $f(x) = 3x$ then we have $|3x-7|$.Now we have to produce a way to be able to tell how much $f(x)$ gets close to $7$ (that is we choose an $\epsilon$), and from here we get how much $x$ has to be close to $2$ so that is true (hence we get $\delta$ which makes it true).
$$0 < |x-2| < \delta \iff -\delta < x-2 < \delta \iff 2-\delta < x < 2+\delta$$
So I write
$$f(2\pm \delta) = 6 \pm 3\delta$$
And then
$$|f(x)-7| < \epsilon \implies |6\pm 3\delta -7| < \epsilon$$
This means $|-1 \pm 3\delta | < \epsilon$.
Now I got stuck because the above expression also reads
$$\frac{1-\epsilon}{3} < \pm \delta < \frac{1+\epsilon}{3}$$
I tried to separate into the two signs of $\delta$ but I don't understand how to conclude that this gives some kind of contradiction, or absurd, or well how this tells the limit is false...
Thank you!