Is this function continuous in its domain?

[Himanshu]

I encountered the following problem in the defination of 'continuity of a function'.

We check the continuity of a function in its domain.

Consider a function f defined by f(x)=(x^2-4)/(x-2).


Its domain is R-{2}. i.e. the continuity of the function will be checked in R-{2}. The function is obviously continuous in its domain. Therefore can we say that the function f is continous.

Or does the function posesses removable discontinuity at x=2.

 

[HallsofIvy]

Homework Statement

I encountered the following problem in the defination of 'continuity of a function' :

DEF- 'Function f is said to be continuous on an interval I if f is continuous at each point x in I.'

Consider a function f defined by f(x)= \frac{x^2-4}{x-2}. Its domain is R-{2}. i.e. the continuity of the function will be checked in R-{2}. The function is obviously continuous in its domain. Therefore can we say that the function f is continous. <br /> <br /> Or does the function posesses removable discontinuity at x=2.

<br /> <br /> What does saying that f is continuous <b>on its domain</b> have to do with &#039;continuous on an interval&#039; which is what the definition you give says. In this case, the domain of the function is not an interval. Yes, this functions has a removable discontinuity at x= 2.

 

[Himanshu]

I have corrected my post. Please have a look at it again.

 

[HallsofIvy]

Yes, it is true that f(x)= \frac{x^2- 4}{x- 2} is "continuous on its domain".

It is not, however, "continuous on the real numbers" nor is it continuous on any interval that includes 2.

 

[Himanshu]

Ok.

If I plot the graph of f it will have a break at x=2. By looking at the graph what do we get to know about the continuity of a function.

I mean to say that if a graph of any function has break points then is the function continuous or non-continous .

 

[ZioX]

Heuristically, a function is said to be continuous on some domain iff you can draw its graph with a pen without lifting the pen from the page.