1. ## lim f(x)

Hi, i was just wondering why f(2) is not defined in the expression below:

f(x) = (x^2 - x - 2) / (x - 2)

I factorised (x^2 - x - 2) to get (x - 2)(x + 1). I then cancelled the (x - 2) in the numerator and the denominator out. The final f(x) would be (x + 1).

However, my calculus textbook says that the function will still be discontinuous at x = 2.

May i ask the principle behind this? Hasn't (x - 2) already been removed from the denominator?

2. Originally Posted by Cplusplusdummy
Hi, i was just wondering why f(2) is not defined in the expression below:

f(x) = (x^2 - x - 2) / (x - 2)

I factorised (x^2 - x - 2) to get (x - 2)(x + 1). I then cancelled the (x - 2) in the numerator and the denominator out. The final f(x) would be (x + 1).

However, my calculus textbook says that the function will still be discontinuous at x = 2.

May i ask the principle behind this? Hasn't (x - 2) already been removed from the denominator?
This is because you are assuming that,
$\frac{x^2 - x - 2}{x-2} = (x+1)$.

This is not true. Because the LHS is not defined at $x=2$.
Remember for two functions to be equal they need to have the same domain.

The correct thing to do is to define,
$f(x) = x+1$ for $x\not = 2$ i.e. leave it undefined.

Then $f(x) = \frac{x^2 - x - 2}{x-2}$ would be correct.

3. Thanks a lot for the explanation! I get it now!