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 $x = 1$.
