Why is this defined even when divided by zero so I've got $$\dfrac{x^3-4x^2+3x}{x^2-1}$$ and want to calculate the asymptotes. There's one a $x=-1$ since the function is not defined there. But the function seems to be defined for $x=1$. How come? 
It should be undefined at $x=1$ since $f(1)=1^2-1 = 0$
 A: Factor the numerator: $x^3 - 4x^2 + 3x = x(x-1)(x-3)$. Cancel the common factor in the numerator and denominator: $$\dfrac{x(x-1)(x - 3)}{(x - 1)(x + 1)} = \dfrac{x(x-3)}{x+1}$$ 
However, please note that the original function is NOT defined at $x = 1$. But no asymptote exists there because it is a removable discontinuity.
Vertical asymptotes occur only when the denominator is zero, but they don't necessarily occur when the denominator is zero. 
In the case $x = -1$, the denominator is zero, so we have a potential asymptote at $x = -1$. Indeed, the numerator of the given function is not zero at $x = -1$, and hence an asymptote does in fact exist, and is given by the vertical line $x = -1$. 
On the other hand, in the case $x = 1$, while the denominator is zero (and hence we need to determine whether or not an asymptote exists there), we see that the numerator also evaluates to $0$ at $x = 1$. Indeed, we see that, both numerator and denominator share a common factor $x - 1$, which can be canceled, and hence the discontinuity at $x = 1$ can be removed. So, in fact, no asymptote exists at $x = 1$.
Suggestion: graph the original function to get an visual idea of what's happening with the given function.
A: The function, as presented, is undefined when $x=1$. However, the discontinuity is removable since the numerator has a factor of $x-1$.
A: It is not defined at $x = 1$; at that point it has a "removable singualarity;" you would show this on a graph by putting a white (hollow) dot at that point.  At a removable singularity, you can redefine the function to make it continuous.  But that changes the function. Conclusion:  Removable singularities are not in the domain of the function.
A: Because there is a factor $(x-1)$ in the numerator.
$$x^3-4x^2+3x=x(x-3)(x-1).$$ So the $(x-1)$ factor in the denominators cancels off.
