Why does the limit as $x$ approaches 1 of $f(x) = \frac{(x^2 - 2x - 3)} { (x- 1)}$ not exist? Why does the limit as $x$ approaches 1 of $f(x) = \frac{(x^2 - 2x - 3)} { (x- 1)}$ not exist?
Can't you use L'Hopital's rule and get $\frac {2x-2}{1} = 0$?
 A: let h defined as $h(x)=x^2-2x-3$, we have $h(1)=-4$
Your numerator tends to-4, while the denominator tends to zero, therefore the limit of your function is +$\infty$ or -$\infty$.
L'hopital rule does not apply to that case
A: No, because you can only apply L'Hopital rule for limits of the form $0/0$, $\infty/\infty$ or $-\infty/-\infty$. The original limit has the form $-4/0$.
A: Note that
$$
f(x)=\frac{x^2-2x-3}{x-1}=\frac{(x-3)(x+1)}{x-1}.
$$
As $x\to 1^+$, $x-1\to 0^+$ and the numerator goes to $-4$. Thus the limit from the right is $-\infty$. Using similar reasoning the limit from the left is $\infty$. Hence the limit does not exist.  
A: Plot your expression on a graphing calculator. Try to find the point where the graph intersects exactly with the line x=1. Not only will you not find it, but you (conventionally) can't even say it approaches infinity. This is because if you "ride along" the slope of the graph, you'll either approach negative or positive infinity depending on which direction you approach x=1.
