Is the function $\frac{x^2-x-2}{x-2}$ continuous? Yesterday,  when  my lecturer  was reading  course lecture, she said that  this function
$$f(x)=\frac{x^2-x-2}{x-2}$$
has a breakpoint and is not continuous, also if we sketch  graph  it will have jumping point   at $x=2$, it is clear from his equation because $x-2$ at $x=2$ has a vertical asymptote, but if we factorize the given equation as $\frac{(x-2)(x+1)}{(x-2)}$  then we could cancel out $x-2$, yes? (as I know in this case $x=2$ is removable point or something like this). If we cancel out, then we are left only with $x+1$, which is continuous everywhere. I have also tried to use wolframalpha to see the graph, here it is

It shows me the graph of straight line, so my question is: was my teacher's belief that  this function is discontinuous  correct or am I wrong?
 A: At $x=2$ it is not defined, and
$$\lim_{x \rightarrow 2} \frac{x^2-x-2}{x-2}=\lim_{x \rightarrow 2} \frac{(1+x)(x-2)}{x-2}=\lim_{x \rightarrow 2} (1+x)=3$$
Because the limit exists, the discontinuity is removable.
A: It is a problem with the language used, continuity is not determined just by the function but also it requires to specify the domain to be specified.
The function is not defined at $x=2$ so it is meaningless to say it is  continuous at 2.
As Daniel pointed out, the discontinuity is removable. But that is something that has nothing to do with the original function, the value of function at x=2 has to be explicitly added.
How ever the function is continuous on $(-\infty , 2) \cup (2,+\infty) $ by default or over any interval not containing 2.
Your lecturer was not wrong , but she was incomplete, she should have mentioned the domain of consideration.
Also you have to understand that the two functions $f_2(x)=\frac{x^2-x-2}{x-2}$ and $f_1(x)={x+1}$ are not the same. To see that at x=2 , $f_1(2)=3$ but $f_2(2)$ is undefined, so the two functions are two different things, their graphs are two similar to see that.
PS : Thanks to DamianSobota for correction.
A: To talk about contiunity of a function  $f$ on a given point $a$ , the function has to be defined on this point and the neighborhood of this point in order to evaluate the limit of $~f~$ when $~x\rightarrow a$ and compare it with $f(a)$ ,  if the limit is equal to $f(a)$ then it's continuous,  otherwise itsn't continuous .
