I still don't get it...
The statement
$\lim\limits_{x \to c} f(x)$ = L
is correct if for any postive number ε there is a δ such that if x is within δ units of c, then f(x) is within ε units of L... ok, I get that.
But then I tried to disprove
$\lim\limits_{x \to 2} \frac{x^2-4}{x-2}$ = 5
And what I got was:
If f(x) is within 0.1 (ε) units of 5 (supposed L) it is either 4.9 or 5.1.
But if I plug in x as 2.9 or 3.1, the resulting f(x) is respectively 4.9 and 5.1; with the δ being 0.9 and 1.1 respectively... If ε is smaller, then I can just get δ closer and closer to 1 to suffice that condition.
And so I still found a δ for every case...
I know this must all sound ridiculous, I'm just hoping to find where I'm thinking wrong.
When you choose a value of $\delta$, you are requiring that every value of $x$ satisfying $0 < |x - x_0| < \delta$ results in a true implication for $|f(x) - L| < \varepsilon$.
In your example, when setting $\varepsilon = 0.1$, you suggest that we can choose $\delta = 0.9$ or $\delta = 1.1$. But that only makes things (sort of) correct at the boundary, which is not enough.
What values of $x$ satisfy $0 < |x - 2| < 0.9$? Not 2.9, not 3.1, but things like 2.1 and 2.5 and 1.4 - and for all of those values of $x$, you'll find that $f(x)$ is more than 0.1 away from 5.