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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

enter image description here

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?

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It is not continuous at $2$ because it is not defined at $2$. But as the picture shows, the discontinuity is removable. The function $g(x)$, defined as above for $x\ne 2$, and as $3$ at $2$, or more simply as $x+1$ everywhere, is continuous. There is no vertical asymptote for $f(x)$. – André Nicolas Nov 26 '11 at 7:37
I can't tell if your teacher said this or you said it, but the function does not have a vertical asymptote at $x=2$. Vertical asymptote refers specifically to the "exploding" behavior you see in a function like $\frac{1}{x}$ near $x = 0$. As your graph demonstrates, this behavior does not take place near $x = 2$. – Austin Mohr Nov 26 '11 at 7:53
vertical asymptote i have added,not my teacher,she said only discontinuity thanks – dato datuashvili Nov 26 '11 at 7:54
A more interesting related question in classical analysis: When in general is a discontinuity for a real valued map removable? A rational function with a factorable polynomial in the numerator with a factor in the denominator is an almost trivial case-but when is it true in general?In complex analysis,for holomorphic functions,there's a very well defined theory based on a theorem of Riemann. But what about real valued maps? Hmmmmmmm............. – Mathemagician1234 Nov 26 '11 at 8:03
it is interesting – dato datuashvili Nov 26 '11 at 8:06
up vote 7 down vote accepted

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.

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thanks @Daniel and all guys – dato datuashvili Nov 26 '11 at 8:26

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.

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You meant: $f_1(x)=x+1$ and so $f_1(2)=3$. – Damian Sobota Nov 28 '11 at 1:46
@DamianSobota : Yes, thank you for correction. – Arjang Nov 28 '11 at 2:15

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