Here is an answer that is more intuitive perhaps but does not require knowledge of calculus.
Consider the nature of your function $f(x)=x^2-1$. You want the range of this function. What is the largest value this function could obtain?
Since $x^2$ will always be positive, we should input the largest number (in terms of "size" or "magnitude") from the domain, namely $x=-2$ (but remember $x\in(-2,1)$; thus, we are not actually using exactly $x=-2$ but a number very close to this). This will yield the maximum value of $x^2$. Thus, $f(-2)=(-2)^2-1=3$ may be thought of as the end point of the range or the largest value obtained for $f(x)$.
As before, $x^2$ will always be positive. Thus, to find the other end point of the range, the minimum value obtained by $f(x)$ will be the input value that minimizes $x^2$, namely $x=0$. Thus, $f(0)=(0)^2-1=-1$. Notice that $x=0$, unlike $x=-2$, is actually in the domain $(-2,1)$. Thus, the range will be $[-1,3)$.
Does that make more sense? It's a pretty "dirty" explanation, but it may be clearer than thinking about it in calculus terms.