# Why isn't there a vertical asymptote at $x = 2$ for $y = \frac{x - 2}{x^2 - 3x +2}$?

Why isn't there a vertical asymptote at $x = 2$ for $y = \frac{x - 2}{x^2 - 3x +2}$?

If you factorize the denominator, you get: $(x-1)(x-2)$. So, when $x = 1$ or $2$, the denominator will be $0$. But I noticed that when I put this function into a graphing program, there is only a vertical asymptote at $x = 1$ and the function seems to be continuous at $x = 2$.

I can also see that the $(x - 2)$ will cancel with the numerator, but I was just wondering on a deeper level why this means that only $x = 1$ is undefined.

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$\displaystyle \lim_{x \to 2}y =$ ? – pedja Apr 30 '12 at 5:13
Let $x\ne 1$, $x\ne 2$. Then $y=\frac{1}{x-1}$. At $x=2$, the function is not defined. The graphing program should really have called the function undefined at $x=2$, but what do programs know? However, if we define $y$ to be $1$ (that is, $\frac{1}{2-1}$) at $x=2$, the function will be continuous at $x=2$. It will not be the same function as the original one, but it is essentially the same. – André Nicolas Apr 30 '12 at 5:17
@pedja: so, it's because the $\lim_{x\to2} y$ exists and $\lim_{x\to1} y$ doesn't? – stariz77 Apr 30 '12 at 5:23
You've got it right: "[T]he function seems to be continuous at $x=2$" in your graphing program. That's because your graphing program isn't making clear the "hole" at $x=2$. The function is in fact undefined there (as it evaluates to "$0/0$"), but doesn't go asymptotic for the reason you have identified: there's an $(x-2)$ in both the numerator and denominator, and these factors cancel each other's tendency to drag the function toward $0$ or $\infty$. (This is an object lesson in not believing everything you (think you) see in a graphing program.) – Blue Apr 30 '12 at 5:25
– pedja Apr 30 '12 at 5:25

We have

$$y = \frac{x-2}{x^2-3x+2}$$

You can check that at $x=1$, we face the following problem

$$y(1) = -\frac{1}{0}$$

So the function is not defined at $x=1$. Similarily

$$y(2) = \frac{0}{0}$$

So the function is not defined at $x=2$ either.

How do we differentiate an asymptote from a removable discontinuity$^{1}$? We need to find the limits. In this case, we have

\begin{align*} \lim_{x\to 2} y &=\lim_{x\to 2}\frac{x-2}{x^2-3x+2}\\ &=\lim_{x\to 2}\frac{x-2}{(x-2)(x-1)}\\ &=\lim_{x\to 2}\frac{1}{x-1}\\ &=1 \end{align*} so there is no asymptote at $x=2$, but rather a "hole". We can fill it in, and in fact make $y$ continuous there by simply defining $y(2)=1$.

However, at $x=1$, as you note, there is an asymptote.

$1:$ [the function] is discontinuous there, but the function can be redefined so that it can be continuous at that point. J.M.

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BTW: Could someone align my equations? I really don't know how. – Pedro Tamaroff Apr 30 '12 at 5:29
OK, this makes sense, but typo in final lim. – stariz77 Apr 30 '12 at 5:35
@stariz77 Fixed. Sorry. – Pedro Tamaroff Apr 30 '12 at 5:36
It wasn't the important part :D – stariz77 Apr 30 '12 at 5:38
@André I like the $=$ in the limits! It is like Landau writes. See page 128 – Pedro Tamaroff Apr 30 '12 at 5:47

The function is not defined at $x = 2$ because of the 0/0 that appears there, but the limit there still exists and so the function does not have an infinity-singularity that would lead to an asymptote.

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The function is:

$$f(x)=\frac{x-2}{x^2-3x+2}$$

Observe that: $$\lim_{x\rightarrow 1^-}{f(x)}=-\frac{1}{0}=-\infty$$ and, $$\lim_{x\rightarrow 1^+}{f(x)}=\frac{1}{0}=\infty$$ And that $$\lim_{x=2}{f(x)}=\frac{0}{0}=(undefined!)$$

We know through L'Hôspital's Rule that:

$$If \lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\frac{0}{0}$$ $$then$$ $$\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\frac{f'(x)}{g'(x)}}$$ $$\lim_{x\rightarrow 2}{\frac{x-2}{x^2-3x+2}}=\lim_{x\rightarrow 2}{\frac{(x-2)'}{(x^2-3x+2)'}}=\lim_{x\rightarrow 2}{\frac{1}{2x-3}}=\frac{1}{2(2)-3}=\frac{1}{4-3}=\frac{1}{1}=1$$

This shows that though $f(2)$ is undefined, the limit is defined. However, for $f(1)$, the limit cannot be rationally defined because $\Big(\big[x\rightarrow 1^{\pm}\big],\big[f(x)\rightarrow \pm\infty\big]\Big)$

Usually, for graphing softwares, $x$-values which are not defined, but have a limit, are graphed as "holes" perhaps missing a pixel or so. These $x$ values are usually either:

1. Values which are not defined on $f(x)$, but have a existing limit.
2. Values which can be computed after "cancelling out" something.

In our case, $f(2)$ is both!

Explaining that there is no asymptote (on graphing software) at $x=2$; if you zoom in enough, the pixels will become messy/non-existant/forming an odd shape (such as a 2-sided-square or circle...), depending on which graphing software you use and its programming.

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