How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:


*

*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.

*Solve $F(x)$ first and then put $x=2$. 
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$
It will give $${F(2)=\frac{0}{4}}$$ which is zero.
How can zero equal not defined?
 A: Remember that functions consist not only of rule of assignment but domain and codomain as well. So, let $$f\colon\Bbb R\setminus\{\pm2\}\to\Bbb R,\quad f(x) = \frac{x^3-4x^2+4x}{x^2-4}$$ and $$g\colon\Bbb R\setminus\{-2\}\to\Bbb R,\quad g(x)=\frac{x(x-2)}{x+2}$$
Just by inspecting domains you can immediately see that these are not equal functions. However, what you did show by your simplification is that restriction of $g$ on $\Bbb R\setminus\{\pm 2\}$ is equal to $f$, i.e. $f(x) = g(x)$ for all $x\neq\pm 2$.
This is rather a common mistake that I believe is due to how algebraic expressions are taught in high school, completely ignoring defining context in which this is allowed. It is allowed, for example, when one takes a limit of the functions:
$$\lim_{x\to 2} f(x) = \lim_{x\to 2} g(x) = g(2) = 0$$
But, $\lim_{x\to 2}f(x) = 0$ does not imply that $f(2) = 0$.
Another trivial example could be functions $f(x) = \frac xx$ and $g(x) = 1$ which are equal at all points but $x=0$. Again, natural domains of $f$ and $g$ are not equal.
A: What you have found is a simplification for $F(x)$, provided $x\neq 2, x\neq -2$.  The denominator $(x^2 - 4)$ of the original function makes it undefined at $x = 2, \;x=-2:$ $(2^2-4) = (-2)^2 - 4 = 0$
So your simplification 
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$ is valid, $\forall x \in \mathbb R \setminus\{-2, 2\}$.
A: The reason is simply that 
$$ \frac{x^3-4x^2+4x}{x^2-4}$$
and 
$$\frac{x(x-2)}{x+2}$$
are two different expressions.
Their values indeed coincide for $x\ne2$ (and they are both undefined for $x=-2$), but they are not "mandated" to be equal at $x=2$.
This symptom reflects the difference between
$$\frac{x-2}{x-2}$$ and $$1.$$
A: "Solve $F(x)$" is the wrong terminology. "Simplify $F(x)$" fits better. One solves problems; one solves equations; one evaluates or sometimes simplifies expressions.
$\dfrac 5 0$ is undefined because there is no number $x$ for which $0x = 5.$
But $\dfrac 0 0$ is undefined because there are many numbers $x$ for which $0x=0$, rather than just one such number.
A basic fact of algebra is that if you plug a number into a polynomial in a variable $x$ and get $0$, then $x$ minus that number is a factor of the polynomial. For example, suppose
$$
f(x) = x^3 -7x^2 + 5x + 21
$$
so that
$$
f(3) = 0.
$$
We conclude that
$$
x^3 -7x^2 + 5x + 21 = (x-3)(\cdots\cdots\cdots).
$$
You still have to do some work to find the other factor, and you get
$$
x^3 -5x^2 + 4x + 6 = (x-3)(x^2 - 4x - 7).
$$
(Factoring $x^2-2x-2$ further will not concern us for now.)
Now suppose you divide this by another polynomial that is $0$ when $x=3$; for example $g(x) = x^2 - 4x + 3.$  Since $g(3)=0$, you conclude that $g(x) = (x-3)(\cdots\cdots\cdots)$, and when you find the other factor you've got $ g(x) = (x-3)(x-1).$
Now look at $\dfrac{f(x)}{g(x)},$ and see that $\dfrac {f(3)}{g(3)} = \dfrac 0 0$ is undefined.
Now simplify:
$$
\frac{f(x)}{g(x)} = \overbrace{ \frac{x^3 - 7x^2 + 5x + 21}{ x^2 - 4x + 3} = \frac{x^2-4x-7}{x-1}}^{\text{when } x\ne3} {} \underbrace{ {} = \frac{-10}{2} = -5}_{\text{when }x=3}.
$$
Are we saying that $\dfrac 0 0 = -5$? Are we saying $\dfrac{f(x)}{g(x)} = -5$ when $x=3$?
No, we're not, because one of the "equals" signs is true when $x\ne3$ and the other when $x=3$, so logically we cannot conclude that $\dfrac{f(3)}{g(3)} = -5.$
However, we can conclude that $\dfrac{f(x)}{g(x)}$ can be made as close to $-5$ as desired by making $x$ close enough, but not equal, to $3$.  And that is expressed by saying $\dfrac{f(x)}{g(x)}$ approaches $-5$ as $x$ approaches $-3$, or by saying $\dfrac{f(x)}{g(x)} \to -5$ as $x\to-3$, or by saying $\lim\limits_{x\to3} \dfrac{f(x)}{g(x)} = -5.$
