<s>[INST] 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?
 [/INST] 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 $$1.$$
</s>