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I am reading Stewarts Calculus book at on page 47. The book gives the following examples $$f(x)= \frac{x^2-x-2}{x-2}$$ and said this function is not continuous at $x=2$. Which I agree at first but then this function is just the line $x+1$ which is continuous at $x=2$. Should I just imagine the function $f$ as a line with a hole at $x=2$ when it is in the rational form? But then it doesnt make any sense to me, when $ \frac{x^2-x-2}{x-2}=x+1$. But the graph of LHS has a hole, and RHS do not. I mean I know LHS is not defined at $x=2$ but they represent the same thing idk something just seems off. Can anyone help me clear this confusion? Thanks

Remu X
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  • When $x-2=0$, you can't divide both the numerator and denominator by $(x-2)$. – peterwhy Jan 03 '24 at 20:56
  • @peterwhy Thank you but technically cant I just cancel that by factoring? So Im just looking at a different algebraic representation of x+1. Why should the graph look different why the algebraic formula are equal? – Remu X Jan 03 '24 at 20:58
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    Yes. Consider $f(x)=\frac{x-2}{x-2}$. It is not defined for $x=2$, although it looks like the constant function $1$. Your example has the same problem. – Dietrich Burde Jan 03 '24 at 20:59
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    You can factorise the numerator $x^2-x-2 = (x+1)(x-2)$. But then to divide the numerator by $(x-2)$? Then $(x-2)$ has to be $\ne 0$. – peterwhy Jan 03 '24 at 21:00
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    The cancellation is invalid when the factor you're canceling is $0$. Otherwise we would have nonsense like $\frac00=\frac{5\times0}0=5.$ So your expression is equal to $x+1$ except when $x=2$, in which case it is undefined. The given function is continuous everywhere it is defined, but it is not defined at $x=2$. – Karl Jan 03 '24 at 21:03
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    I see, so I should really think the equality holds except when $x=2$. Thanks – Remu X Jan 03 '24 at 21:04

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