Are these two functions equal? If $f(x) = \frac{x^2 − 4}{x − 2}$ and $g(x) = x + 2$, then we can say the functions
$f = g$
I think this is false but I cannot prove it.
 A: The functions are not equal because they have different domains.
A: $f(2)$ is undefined and $g(2)=4$, so they are different functions.
However for $x\neq2$, $\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$. So on the restricted domain $(-\infty,2)\cup(2,\infty)$, they are the same function.
A: Some teachers might tell you these functions are the same. Other teachers will tell you that they are not. Both viewpoints have merit. However, usually those espousing equality are just being sloppy.
$$f(x) = \dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2=g(x)$$
The above statement is true everywhere it is defined (i.e. as long as $x \not= 2$). 
In a loose sense $f$ and $g$ are the "same". 
However, strictly speaking the function $f$ and the function $g$ are equal on the domain of $f$. Yet they are different functions since $g$ is defined at $x=2$ while $f$ is not (they have different domains).
Now for many purposes these two functions can be treated as equal (which is why some textbooks/teachers will say they are the "same"). For example:
$$ \lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} \dfrac{x^2-4}{x-2} = \lim\limits_{x \to 2} \dfrac{(x-2)(x+2)}{x-2} = \lim\limits_{x \to 2} x+2 = \lim\limits_{x \to 2} g(x) = g(2) = 2+2=4$$
The reason we can do this with the limit is that the limit concerns points "close to" $x=2$. However, $x$ is never allowed to be exactly equal to $2$ (in the limit). Since we aren't actually working at $x=2$, $f(x)=g(x)$ everywhere we are limiting. :)
A: In this case you can replace $x$ with a known value in the domain of both functions.
Remember you're testing for $f(x)=g(x)$ so for the statement to be true the result after solving for one value of $f$ (say $f(2)$) is going to have to be the same for $g$.
To make it clear:
IF $f(x) =g(x)$ is true then:
$f(1) = g(1)$
$f(2) = g(2)$
etc...
Hope it helps.
