# 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!

I do that as the numerator can not be factorized.

This is incorrect. The numerator can be factorized as: $(x^2 - 1) = (x-1)(x+1)$

Now, $y = \frac{(x-1)(x+1)}{x(x-1)}$. As you correctly pointed out, the denominator becomes $0$ at $x = 0$ and $x = 1$. Therefore, the domain of the function is $(\infty, 0) \cup (0, 1) \cup (1, \infty)$.

Now, consider the case when $x \ne 1$. The function $y = \frac{(x-1)(x+1)}{x(x-1)} = \frac{x+1}{x} = 1 + \frac{1}{x}$ when $x \ne 1$.

Therefore, $y - 1 = \frac{1}{x}$. Let $g(x) = \frac{1}{x}$. As you can see, at $x = 0$, you get the $1/0$ form. But we already know that $0$ is not in the domain of $g(x)$ (and $f(x)$).

Next, let us examine the vertical asymptotes of $f(x)$. As $x \to 0^+$, $f(x) \to +\infty$. As $x \to 0^-$, $f(x) \to -\infty$. Therefore, the vertical asymptotes are in different directions on different sides of $0$. For horizontal asymptotes, as $x \to +\infty$, $f(x) \to 1$. Similarly, as $x \to -\infty$, $f(x) \to 1$.

Now let us examine the discontinuity in $f(x)$ at $x = 1$. $\lim\limits_{x \to 1^+}(1 + \frac{1}{x}) = \lim\limits_{x \to 1^-}(1 + \frac{1}{x}) = 1$. Note that we took the simplified expression of $f(x)$ because $x \ne 1$, we are only taking the limits of $x$ approaching $1$ from both sides. Hence, the function is "smooth" around $x = 1$ except for a discontinuity at $x=1$ (function is undefined at $x = 1$).

Now, can you plot the function? You have already seen that it is the same as $1 + \frac{1}{x}$ except for being undefined at $x = 1$.

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.