Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same? Let $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$.

It is clear that $f(x)=g(x)$ for all real numbers $x$.

My question: Why do $f$ and $g$ not have the same domain of definition, however $f(x)=g(x)$?

Note: I know only that $f$ extended by continuity at $x=1$ as a reason, I hope to know more reasons if it is possible.
The two functions are not the same because $f(x)$ is not defined for $x=1$, so its graph is a stright line that has a ''hole''. 



The Domain of definition is the subset of $\mathbb{R}$ where the function is defined and this means that has values in the codomain. Division by $0$ is not defined so the function $f(x)$ cannot be defined for $x=1$. You can extend it by continuity but this means that you have a new function $f_1(x)$ that coincide with $f(x)$ for $x \ne 1$ and is defined also in this point in such a way that 
$$
f_1(1)=\lim_{x\rightarrow 1} f(x)
$$
and, in this case, you really have $f_1 =g$.
I personally think it's silly to say that $ f(x) = \frac{x^{2}-1}{x+1} $ at $x = 1$ is undefined. I've been giving this some thought lately. It's like saying that $ f(x) = x$ is undefined at $x = 0$ just because it's equal to $\frac{x^{2}}{x}$ (multiply $\frac{x}{1}$ by $\frac{x}{x}$; if you don't agree that they're equal then you also don't agree that $x \times 1 = x$).
I completely agree that $\frac{1}{x}$ at $x = 0$ is undefined, but this is a different scanario. Just look at the graphs:


With $\frac{1}{x}$, the line completely swerves off into positive and negative infinity. $\frac{x^2}{x}$ looks completely identical to $f(x) = x$. With $\frac{x^2}{x}$ you can approach $0$ from both directions and both limits will be $0$. The notion of discontinuity doesn't seem particularly useful in this case.
Just my opinion, though. Clearly $\frac{x^2}{0}$ is undefined, which definitely indicates something strange about notation.