Why can I remove discontinuities by modifying function Given for example $$\lim_{x\to 2} \frac{x^{2} + x - 6}{x - 2}.$$ The limit cannot be computed in that form of the function.
But WHY am I able to modify the expression so as to define the limit as $x$ approaches $2$? What is happening? Please ELI$5$ how that is valid math. 
 A: We do not "define" the limit to be the limit of the "modified" expression. The limit of the "modified" expression is equal to the original limit. 
It is better to see why this makes sense in the simple cases. For instance,
$$\lim_{x \to 0} \frac{x}{x}$$
The limit cannot be computed in that form for the same reason your limit can't. However, whenever you pick a very small non-zero value of $x$ and compute $x/x$, the result is approximately (actually, exactly) $1$. This is what it means for $\lim_{x \to a} f(x)$ to be equal to $l$, where $f$ is a function defined in a neighborhood of (i.e. in a sufficiently small region around) $a$ (but not necessarily defined at $a$):

For any $\epsilon > 0$, there exists a $\delta > 0$, such that if $0 < |x - a| < \delta$, then $|f(x) - l| < \epsilon$

This means that however small you want the distance between $f(x)$ and $l$ to be, you can find a sufficiently small number $\delta$, such that for any $x \neq a$  within $\delta$ from $a$, $f(x)$ is within $\epsilon$ from $l$.
So, when we cancel out the $x$'s in the above limit, we are doing perfectly valid math. We do not care about the value of the function at the point $0$ (which is undefined), but about the value of the function in a neighborhood of $0$ which does not include $0$. The same reasoning applies to any such limit as yours.
A: $\lim_{x\to 2} \frac{x^2 + x - 6}{x - 2}$
The limit absolutely can be calculated in this form.  Go back to the definition of a limit.
$\lim_\limits {x\to a} f(x) = L$ means:
$\forall \epsilon > 0, \exists \delta >0 \text{ such that, } 0<|x-a|<\delta\implies |f(x) - L| < \epsilon$
Find $\delta$ such that: $0<|x-2|<\delta\implies | \frac{x^2 + x - 6}{x - 2} - 5| < \epsilon$
$|\frac{x^2 -4x + 4}{x - 2}| < \epsilon\\
|\frac{(x -2)^2}{x-2}| < \delta$
When $\delta<\epsilon, | \frac{x^2 + x - 6}{x - 2} - 5| < \epsilon$  
Well what does that mean?  When $x$ is in the neighborhood of $2, f(x)$ is in the neighborhood of $5.$
Neighborhood? that sounds a little fuzzy.  It makes is sound like $f(x)$ is only kinda-sorta close to $5.$  But $\epsilon$ is the radius of this neighborhood, and we can make $\epsilon$ as small as we want to.  So, it really isn't fuzzy at all.
A: There is only a problem when $x = 2$. I'm sure you agree there. However, when we are looking at a limit $\lim_{x\to a} f(x)$ we are not considering the case where $x = a$, only when $x$ is close to $a$. So assume that $x\neq 2$, then
$$\require{cancel}\frac{x^2+x-6}{x-2} = \frac{\cancel{(x-2)}(x+3)}{\cancel{x-2}} = x+3  $$
and hence the limit approaches $2+3 = 5.$
A: $$\lim_{x\to 2}\frac{(x-2)(x+3)}{x-2}$$
The above is $x+3$ for $x\ne 2$. Thus, the limit at $x\to 2$ is $5$. 
A: You can define the limit because the function has a "removable discontinuity". That is to say, the function is similar to the $x + 3$ function over its entire domain, except for the point $x = 2$. At that domain value, the function isn't defined.
In other words, your function is formed by taking $x + 3$, and removing a single point to create an infinitesimally narrow discontinuity.
To make that function continuous, all we have to do is declare that its value is 5 at $x = 2$, restoring the only point that is missing. In all other regards, it is identical to the function $x + 3$.
The limit exists because the limit, by definition, involves an imaginary process of approaching a domain point of a function arbitrarily closely, from above or below, without actually reaching that point. In this situation, we can approach $x = 2$ from either above or below. As we approach more and more closely to $2$, the value of the function gets closer and closer to $5$. No matter how close we get, there is room to get closer. Of course, the $5$ isn't actually there, and we must not step on $x = 2$ because then we have "fallen off the function". That's the intuitive gist of the limit.
By the way, when we have a fraction of polynomials, we can apply long division to see whether they "evenly divide". We don't have to perform any factoring:
              1    3  <-  i.e.  x + 3
         ____________
  1 -2  | 1   1   -6  <-  i.e.  (x^2 + x - 6) / (x - 2)
          1  -2    :  <-  i.e.  x - 2 goes into x + 1 once, remainder 3
         ------    :  <-  bring down the 6
              3   -6
              3   -6  <-  x - 2 goes 3x - 6   3 times
              ------
                   0  <- No remainder!

If you go through this long division and there is a remainder, then the singularity is not removable. Let's try, say $\frac{x^2 + 5x + 1}{x + 3}$:
              1   2
         ___________
  1  3  | 1   5   1
          1   3
          -----
              2   1
              2   6
              ------
                 -5

We have a remainder of -5. What does this mean? This:
$$\frac{x^2 + 5x + 1}{x + 3} = (x + 2) - \frac{5}{x + 3}$$
Basically, the graph of this function looks a lot like $x + 3$ far away from the $x = -3$ singularity.  Far away from this point, the residue term $\frac{-5}{x + 3}$ approaches zero and contributes little to the function.
But near $x = -3$ it doesn't look that way: the residue term forces it to curves off into negative infinity as it approaches $-3$ from below, and toward positive infinity as it approaches $-3$ from above, and thus doesn't approach any limit. The discontinuity is not removable, and the function not only has no value at $x = -3$, but also no limit.
So you see, for a fraction of polynomials, it's a special case when you can remove the discontinuity: it's only when they divide evenly with no remainder.
