How is $\frac {x^2 + x - 6}{x - 2}$ different from $(x+3)$? Let's consider the function: 
$$ f(x) = \frac {x^2 + x - 6}{x - 2} $$
At x = 2, the value fo the function is undefined, because $ x - 2 $ = 0. But if factor the expression on the numerator, we get $ \frac {(x + 3) (x-2)}{x-2} $, which on simplifying yields $ x + 3 $. At $ x = 2 $, the function gives 5, which is not what we get before simplifying. 
If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?
 A: The simplification is only valid for $x\neq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.
In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $x\neq 2$”.
It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.
A: They do agree. But remember that $\frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.
A: Note, that when a function is defined, it's not only about a formula but also about its domain.
For the function given
$$f(x) = \frac {x^2 + x - 6}{x - 2}$$
if it is to be expressed in that way, it must clearly be $x\neq 2$ and thus its domain is $D_f = \mathbb R \setminus \{2\}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.
A: This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point. 
Take this as another example: 
$$f_1(x) = \frac{x(x+1)}{x+1} \quad f_2(x) = x$$
At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x \in \mathbb{R}$ whereas $f_1(x)$ is defined for $x \neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.
A: Surely not because you are manipulating the definition of continuity. For example$$f(x)=\begin{cases}1\quad,\quad  x\ne 0\end{cases}\\g(x)=1$$ are clearly different. One is continuous on $\Bbb R$ and one is not but we have $$f(x)={x\over x}\ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.
