Are there general guidelines to make "assumptions" when proving limits? I am studying the definition of the limit using Paul's Online Notes
When proving the following limit (Example 3)
$\lim\limits_{x \to 2} x^2+x-11 = 9$
At one point he assumes:
$|x+5| < K$
He also asumes:
$|x-4| < 1$
Because one is a nice number to work with.
I have read multiple explanations and every author seems to make their own "assumptions" when dealing with limit proofs. So my question is:
Why the assumptions they make are right?
Lets say I have to prove the following:
$\lim\limits_{x \to 4} x = 2$
While the above is impossible, what stops me from making my own "assumptions" like 2=4 to prove the limit?
Thanks in advance.
 A: You may not make absurd assumptions such as $2=4$.  However, making assumptions about $x$ is perfectly legal so long as it coincides with we want to prove.
When dealing with limits, as in Paul's example, we can restrict our attention only to neighborhoods close to our point of interest.
Take for an exaggerated example $\lim\limits_{x\to 5} f(x)$ where $f(x)=\begin{cases} \text{something overly complicated and ugly to work with}&x\geq 6\\ 2 & 4\leq x<6\\ \text{something else incredibly complicated}&x<4\end{cases}$
If I try to analyze the function outside of the range $[4,6)$, then I will have an incredibly difficult time going about it.  However, it doesn't matter at all what is going on outside of that range since we are asking about finding the limit $\lim\limits_{x\to 5}f(x)$.  By restricting our attention to when $x$ is within a neighborhood of $5$ (since we only care about what happens in a neighborhood of $5$), we see that whenever $|x-5|<\delta~~~(0<\delta\leq 1$ in this case $)$, in otherwords, $x\in (5-\delta, 5+\delta)$, you fall entirely within the easier case of the piecewise defined function that I defined above, and the function is incredibly easy to work with.  Within that neighborhood, the function is equal to precisely 2.  It follows that in my example $\lim\limits_{x\to 5} f(x) = 2$ since whenever $|x-5|<\delta<1$ you have $|f(x)-2|=|2-2|=0<\epsilon$ for every $\epsilon>0$.
In much the same way, Paul has decided to use the fact that he only cares about the properties of his function near the point in question (in his example $\lim\limits_{x\to 4} x^2+x-11$), so by restricting his attention to a small enough neighborhood around the point, he could understand the structure of the function more easily than otherwise.

A friendly reminder, in proving limits with an epsilon-delta argument:

$\lim\limits_{x\to c} f(x) = L$ if and only if for every $\epsilon$, there exists a $\delta$ such that whenever $0<|x-c|<\delta$ it follows that $|f(x)-L|<\epsilon$

