Why aren't these functions the same?

  1. I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2

    Now everyone would agree that f has a domain R\{1} and g has a domain R.

    Yet I can write (x^2+x-2)/(x-1) = x+2

    So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?
     
  2. jcsd
  3. Mentallic

    Mentallic 3,648
    Homework Helper

    If you evaluate f(1) then you get an undefined value 0/0. Of course we know that

    [tex]\lim_{x\to 1}f(x) = 3[/tex]

    But just because the limit exists doesn't mean that the function is defined at that point.
     
  4. I understand that. But why can't I write g(x) = (x^2+x-2)/(x-1) = x+2 ?
     
  5. pwsnafu

    pwsnafu 902
    Science Advisor

    Only if x is not equal to 1. So you need to write ##\forall x \neq 1, \,\frac{x^2+x-2}{x-1} = x+2##

    But you didn't rewrite the expression. If I define
    ##h : \mathbb{R} \to \mathbb{R}## with ##h(x) = \frac{x^2+x-2}{x-1}## and ##h(1) = 3## then that is indeed equal to g(x), but not equal to f(x).

    There is a difference between you can do something, and you did something.
     
    Last edited: Apr 30, 2013
  6. micromass

    micromass 18,442
    Staff Emeritus
    Science Advisor

    I wouldn't agree with this. The domain is something chosen by the person who defines it. All we can say is that the maximal possible domain of ##f## is ##\mathbb{R}\setminus \{1\}##. But the domain can possibly be much smaller if we choose it to be.

    The equation

    [tex]\frac{x^2 + x - 2}{x-1} = x+2[/tex]

    is only valid for ##x\in \mathbb{R}## with ##x\neq 1##. For ##x=1##, it is not true. So we have that ##f(x) = x+2## for all ##x\in \mathbb{R}\setminus \{1\}##. The value ##f(1)## still isn't defined.

    That ##f(1)=3## somehow, is false. However, this is why limits are invented. So we can say that
    [tex]\lim_{x\rightarrow 1} f(x) = 3[/tex]

    So although ##f(1)## doesn't make sense, we can take the limit. The limit denotes the value that ##f(1)## would have been if it were defined in ##1## and if ##f## were to be continuous.
     
  7. pwsnafu

    pwsnafu 902
    Science Advisor

    In fact it could also be larger. No body said x couldn't be complex.
     
  8. micromass

    micromass 18,442
    Staff Emeritus
    Science Advisor

    Very true!
     
  9. So if I previously define the domain, I can't change that domain unless I write an entirely new function?

    Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h?
     
  10. micromass

    micromass 18,442
    Staff Emeritus
    Science Advisor

    Yes to both.
     
  11. Cool, thanks for helping me clear my confusion. I guess that never really got explained to me by anyone and I never picked up on it.
     
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