Does the function
$R(x) = \frac{x^2  x  2}{x  2}$
have a vertical asymptote at $\,x = 2\,$?
No!
Some wellintentioned web sites conclude that this function has a vertical asymptote
at $\,x = 2\,,$
simply because the denominator is zero there.
This is incorrect.
In order to find a vertical asymptote,
it is NOT ENOUGH to just set the denominator equal to zero.
You must ALSO check that the numerator is nonzero!
For this particular function, substitution of $\,x = 2\,$ into the numerator gives:
$$\cssId{s14}{2^2  2  2\ \ = \ \ 4  2  2 \ \ = \ \ 0}$$
Therefore, both the numerator and denominator are zero when $\,x = 2\,.$
When BOTH the numerator and denominator of a rational function are zero,
then you must work harder to figure out what's going on.
You must determine what's happening CLOSE TO the $x$value of interest.
Further analysis shows:
$$
\cssId{s19}{R(x)}
\cssId{s20}{= \frac{x^2x2}{x2}}
\cssId{s21}{= \frac{(x+1)(x2)}{x2}}
\cssId{s22}{= (x+1)\cdot\frac{x2}{x2}}
$$
Cancelling off the extra factor of one:
$$\cssId{s24}{R(x) = x + 1\,,}\
\cssId{s25}{\text{providing } x\ne 2}$$
The function $\,R\,$ isn't defined when $\,x = 2\,,$ since division by zero isn't allowed.
For all values of $\,x\ne 2\,,$ ‘$\,\frac{x2}{x2}\,$’ is just the number $\,1\,$!
The graph of $\,R\,$ is therefore the same as the line $\,y = x + 1\,,$
except that it has been ‘punctured’ at $\,x = 2\,.$
From this renaming, it is now clear that when $\,x\,$ is close to $\,2\,,$ the outputs are close to $\,2 + 1 = 3\,.$
The graph of the function $\,R\,$ is shown below:
DEFINITION
puncture points (holes)
Let $\displaystyle\,R(x) = \frac{N(x)}{D(x)}\,$ be a function.
The function $\,R\,$ has a puncture point (hole) at $\,x = c\,$
if and only if
the following three conditions hold:
 $D(c) = 0\,$ (the denominator is zero at $\,c\,$)
 $N(c) = 0\,$ (the numerator is zero at $\,c\,$)
 as $\,x\,$ approaches $\,c\,,$ the values $\,R(x)\,$ approach a finite number
NOTES ABOUT PUNCTURE POINTS (HOLES):

THIS DEFINITION HOLDS FOR RATIONAL FUNCTIONS, AND MORE!
It might be that both $\,N\,$ and $\,D\,$ are polynomials, making this a rational function (i.e.,
a ratio of polynomials).
However, $\,N\,$ and $\,D\,$ could be more general functions.
For example, a bit of calculus shows that the function $\displaystyle\,R(x) = \frac{{\text{e}}^{x2}  1}{x  2}\,$ has a
hole at $\,x = 2\,.$
Note:
In calculus, a ‘hole’ is an example of a removable discontinuity.

YOU NEED ALL THREE CONDITIONS:
It it not enough merely to require that both the numerator and denominator are zero.
Why?
Instead of a puncture point, you might have a vertical asymptote.
For example, the function $\displaystyle\,R(x) = \frac{x2}{x^2  4x + 4} = \frac{x2}{(x2)(x2)}\,$ has
a vertical asymptote at $\,x = 2\,.$
Why?
For all $\,x\ne 2\,,$ $\displaystyle\,R(x) = \frac{1}{x2}\,.$

TYPICAL APPROACH FOR ANALYZING A RATIONAL FUNCTION FOR VERTICAL ASYMPTOTES/HOLES:
 Set the denominator equal to zero.
 Check the resulting value(s) in the numerator.
 If the denominator is zero and the numerator is nonzero, then you're guaranteed to have a vertical asymptote.

If the numerator and denominator are BOTH zero, then you must do some additional analysis:
you have either a vertical asymptote or a hole.