# Why $(x^3-8)/(x-2)$ is not defined at $x=2$?? [duplicate]

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.

• The expression $g(x) = (x^3-8)/(x-2) = (x-2)(x^2+2x+4)/(x-2)$ is not defined at $x=2$. Cancelling the $(x-2)$ is an invalid operation as you are effectively dividing numerator and denominator by zero – Mufasa Sep 30 '15 at 23:36
• – Blue Sep 30 '15 at 23:37

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".

• I'm a mathematician who would not say that $g$ is defined at $2$. – Michael Joyce Oct 1 '15 at 1:29
• @MichaelJoyce Noted. – Callus - Reinstate Monica Oct 1 '15 at 3:08

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:

• What do you mean it's not continuous at $x=2$? If you define $g(2) = 12$ then it is continuous – Callus - Reinstate Monica Oct 1 '15 at 0:21
• In fact the term to use is probably is that the function has a "removable discontinuity" at $x=2$ as in here:wolframalpha.com/input/?i=discontinuities+%28x^3-8%29%2F%28x-2%29 – NoChance Oct 1 '15 at 0:35
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.