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:
- Values which are not defined on $f(x)$, but have a existing limit.
- 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.