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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

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    $\begingroup$ 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 $\endgroup$ – Mufasa Sep 30 '15 at 23:36
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    $\begingroup$ Related: Why does factoring eliminate a hole in the limit? $\endgroup$ – Blue Sep 30 '15 at 23:37
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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".

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

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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:

Link1

Link2

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  • $\begingroup$ What do you mean it's not continuous at $x=2$? If you define $g(2) = 12$ then it is continuous $\endgroup$ – Callus - Reinstate Monica Oct 1 '15 at 0:21
  • $\begingroup$ The function has no value at x=2 as is, but it has a limit as x approaches 2. So the two functions have the same limit but they are not equal. Maybe I should phrase this better. $\endgroup$ – NoChance Oct 1 '15 at 0:27
  • $\begingroup$ 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 $\endgroup$ – NoChance Oct 1 '15 at 0:35
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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.

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