The problem:
$$g(x) = \frac{x^3-8}{x-2}$$
Explain why $g$ is not defined at $x=2$.
My solution:
$$g(x) = \frac{x^3-8}{x-2} = \frac{(x-2)(x^2+2x+4)}{x-2}$$
The two $(x-2)$ get canceled $\implies x^2 + 2x + 4$.
--> This should be undefined at $x=2$, not really.
But if we don't change the function it is of course not defined at $x = 2$, because it would give us a $0$ (zero) at the denominator. That gives us infinity.
Please help, not if sure if I am on the right track. Thank you very much
you have understood the situation well. The formula for $g$ does not give a value at $x=2$. The formula is just undefined. However, there's a unique way to extend $g$ to be continuous at $x=2$, which is by setting $g(2)=12$, as you noted. One quibble, though, is that having a $0$ in the denominator does not "give us infinity".
Language note: Most mathematicians ( in my experience ) would say that $g$ is defined at $2$. While that's not quite correct, it's kind of a short hand for the above "unique extension".
The domain of the division operator on $\mathbb{R}$ is $$\{(a,b)\in\mathbb{R}^2 | b\not=0\}.$$ Using the rules of composition, you can see that $$x\mapsto {x^3 - 8\over x - 2 }$$ is not defined at $x = 2$.
The problem is that this function:
$$g(x) = \frac{x^3-8}{x-2}$$
does not have a value at $x=2$ because it is not continuous at $x=2$ (or you could say it has a removable discontinuity at $x=2$) as a result,
$$g(x) = \frac{x^3-8}{x-2} = \frac{(x-2)(x^2+2x+4)}{x-2}$$
is NOT true for $x=2$
It is NOT correct to say "The two $(x-2)$ get canceled" because we can't divide by $(x-2)$ when $x=2$.
The two functions above have different domains but the same limit value at $x=2$.
Some what a similar argument arises when some one says that if $\frac{1}{x-2}=\frac{2}{x-2}$, cancel the (x-2) out to get $1=2$, again we can't cancel (x-2) out in the denominator when it is zero.
A graph for a function with point of discontinuity represent the y value with a small open circle. See for example:
WolfarmAlpha plot of the function with a hole representing discontinuity.
The following links talk about the same issue you have here:
You can only factor $ x -2$ in the numerator and make the division if $x\neq 2$. The function is not defined in $x=2$ because the division is not defined in this point.
