As an example: $$ F(x) = \frac{(1-x)^2-1}{(1+x)^2-1} $$
When $x=0$ this involves $0/0$ and calculators output 'undefined'.
But when looking at a graph of the function it is intuitively clear that the answer should be $F(0) = -1$.
Is there something that can be done with the function to get the expected output?
Using that
$$A^2-B^2=(A-B)(A+B)$$
We have
$$F(x) = \frac{(1-x)^2-1}{(1+x)^2-1} = \frac{(1-x)-1}{(1+x)-1}\cdot \frac{(1-x)+1}{(1+x)+1}=$$ $$ \frac{-x}{x}\cdot \frac{(1-x)+1}{(1+x)+1}=-1\cdot \frac{2-x}{2+x}\to -1$$
indeed then recall that we can cancel out the $x$ terms and take the limit.
Refer also to the related:
One never "guesses" such an answer. Never.
Instead, compute
$$\lim\limits_{x \to 0} {(1-x)^2 -1 \over (1+x)^2 -1}$$
which is undefined, so you use l'Hospital's rule, which gives the answer you expect.
