Another about limits, vertical asymptote I am asked to find the vertical asymptotes if any of the following rational function:
$$\begin{align} y= (x^2-1)/(x^2-x)\end{align}$$ so what I do is first of all is to find the domain of the function, so $$\begin{align}x^2-x=0\end{align}$$ to find the solution to that I factorize the expression and I get $$\begin{align}x(x-1)\end{align}$$ so the values I get are $0$ and $1$, now what I would do is to plug the values $0$ and $1$ into the function, I do that as the numerator can not be factorized. by putting $1$ I get $0/0$ and by putting $0$ I get $k/0$, so I would say that only $0$ is a vertical asymptote, so my question is the way I am proceeding is it the right one to find vertical asymptotes if any? i.e


*

*looking for domain of the function

*factorize if possible the numerator

*plug the results given in step 1 into the rational function and I will only get rational asymptote in those values which give me a
rational function of the form $k/0$ and I discard those ones, as
vertical asymptotes, which give me a a rational function of the form
$0/0$


Is that right?
Thanks!

Not exactly. What you are trying to do is remove common factors from the numerator and denominator. In the example above, the common factor is $x-1$ (in other words, the function has a removable singularity at $x=1$).
If the numerator and denominator both evaluate to zero at a particular point, you could use l'Hôpital's rule (possibly repeatedly) to see if the common factors cancel. In the above example, applying l'Hôpital's rule at $x=1$ results in $\frac{2}{1}$, so there is no vertical asymptote at this point.
However, with the rational function $x \mapsto \frac{x^2-1}{(x^2-x)(x-1)}$, you can see that both the numerator and denominator both evaluate to zero at $x=1$, but the function does have a vertical asymptote at $x=1$. Using l'Hôpital's rule in this instance would result in a numerator of $2$ and a denominator of $0$ which indicates a pole at $x=1$ that is not cancelled. This illustrates that if the numerator and denominator evaluate to zero, then you need to investigate further to determine whether or not there is an asymptote at that point.
