Why must polynomial division be done prior to taking the limit? Suppose I wish to evaluate the following,
$$\mathop {\lim }\limits_{x \to 2} \left( {{{{x^2} - 4} \over {x - 2}}} \right)$$
If I just substitute two into $x$, it can't be done because the answer would be undefined (division by zero).
But, if I complete the polynomial division, that I hate to do because I'm all thumbs at it,
$$\mathop {\lim }\limits_{x \to 2} \left( {{{{x^2} - 4} \over {x - 2}}} \right) = \mathop {\lim }\limits_{x \to 2} \left( {x + 2} \right) = 4$$
Please tell me what's going on here?

The point is that the functions
$$\frac{x^2-4}{x-2}\quad\hbox{and}\quad x+2$$
are equal except at $x=2$, where the second is defined and the first is not.  If you look closely at the definition of a limit as $x\to a$, you will see that it is carefully framed in such a way that the value of the function (if any) when $x=a$ is irrelevant.  Therefore the two functions above have the same limit as $x\to2$.  However, as you have noted, you cannot just substitute $x=2$ in the first as it is undefined.  On the other hand, the second function is defined at $x=2$, and better still, it is continuous at $x=2$, because it is a polynomial.  Therefore, using the definition of continuity,
$$\lim_{x\to2}(x+2)=2+2=4\ .$$
And finally, as already noted,
$$\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4\ .$$
This kind of problem is superficially very simple, but as you can see, there is quite a lot behind it if you want to fully understand what is going on.
