Limit Different if I Factor and Simplify an Equation First. So I have $$\lim_{t\to -1} \frac {t^2 - t - 2}{t^2 - 1} = \frac 32$$ The solution given was found by factoring and simplifying the equation and then taking the limit of numerator and denominator and then the individual terms. 
I tried do this without factoring and simplifying. This meant I just took the the limit of the numerator and denominator and then each individual term. This gave me $\lim_{t \to -1} \frac{t^2 - t -2}{t^2 - 1} = \frac{1-1-2}{1-1}$. This gives me a 0 in the denominator so it can't be used. 
My question is why my method didn't work and why I need to factor and simplify first. I understand it makes things easier to do that first but shouldn't it still have worked?  
 A: You should have gotten $$\frac{1 - (-1) - 2}{1-1} \leadsto \frac 00$$
This is an indeterminate form, which simply indicates that more work needs to be done, e.g., factoring and canceling the common term in the numerator and denominator, or performing L'Hospital.
Note that if you are taking the limit of a rational function $\lim_{x \to a} \frac{f(x)}{g(x)}$, where $a$ is a zero of both $f(x)$ and $g(x)$, then $(x - a)$ is a common factor of both $f(x)$ and $g(x)$, the simplest route is likely to be canceling that common factor. Recall that $\frac{f(x)}{g(x)}$ need not be defined at $a$ when taking limits; what is important is what is happening as $x$ approaches $a$, not what happens exactly at $a$.
A: Since both numerator and denominator are polynomials with root $-1$, you should be able to get the factor $(x+1)$ there. 
More generally: If $c$ is a root of $P(x)$, then $P(x)=(x-c)Q(x)$, where $Q(x)$ is a polynomial. You can obtain this polynomial using long division or Horner scheme.
So try to rewrite both polynomials in this way.
You should get:

 $$\frac{t^2 - t -2}{t^2 - 1} = \frac{(t+1)(t-2)}{(t+1)(t-1)} = \frac{t-2}{t-1}$$

You also asked:

Why my method didn't work and why I need to factor and simplify first.

Well, the reason why simply plugging in the value $t=-1$ into the expression does not work in this case is simply that the given expression is undefined for $-1$. (Since you get $\frac00$.)
