How to disprove a false limit without knowing the limit I was thinking about the definition of limit, and I'm trying to figure out how to disprove instead of proving a limit. For instance, if I define the $\lim_{x\to1} \frac{x^2-1}{x-1}$ as 3. Could we find a contradiction in that definition without proving that the limit is actually 2?
 A: Sure. So, the definition of the limit of a function at a point has, within it, a way to disprove that a given number is not the limit at that point.
More precisely, let $f(x)$ be a function that is defined in some deleted neighborhood of a point $a \in \mathbb{R}$. Saying that $\lim_{x \to a} f(x) = L$ is equivalent to saying the following statement:
$$\forall \epsilon >0: \exists \delta > 0: \forall x \in \text{domain}(f): 0 < |x-a| < \delta \implies |f(x)-L| < \epsilon$$
It follows that $\lim_{x \to a} f(x) \neq L$ iff:
$$\exists \epsilon > 0: \forall \delta > 0: \exists x \in \text{domain}(f): 0 < |x-a| < \delta \land |f(x)-L| \geq \epsilon$$
So, in principle, if you want to show that $f$ does not tend to the number $L$ as $x \to a$, you need to show that there is an $\epsilon > 0$ such that $|f(x)-L|$ doesn't get arbitrarily small even when you get close to $a$. This doesn't require you to show that $f$ tends to an entirely differently limit and that should make sense because functions don't need to have limits at every point.
A: Since taking the limit $\;x\to1\;$ means $\;x\;$ is arbitrarily close to $\;1\;$ but always different $\;1\;$ , we have that as $\;\frac{x^2-1}{x-1}=x+1\;$, and then
$$\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}(x+1)$$
and assuming the above limit is $\;3\;$, we'd get:
$$\left|\frac{x^2-1}{x-1}-3\right|=\left|(x+1)-3\right|=|x-2|\;\;(**)$$
Now we can chose $\;|x-1|<\frac12\iff -\frac12<x-1<\frac12\iff-\frac32<x-2<-\frac12\;$ , so that
$\;x-2<0\;$ and thus
$$\;|x-2|=-x+2>\frac12$$
This means that $\;|x-2|>\frac12\;$ , so choosing $\;\epsilon =\frac14\;$ causes a straightforward contradiction to
the definition of limit of function.
