Re: How do you find the limits, holes, and vertical asymptotes of rational functions?
Originally Posted by david213
f(x)=x^2+4/x^2-5x-6
Given $\displaystyle \frac{x^2+4}{x^2-5x-6}$.
There are no zeros for the numerator.
Both $\displaystyle -1~\&~6$ are zeros for the denominator.
Start there.