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Continuity of a function

  1. Aug 29, 2007 #1
    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 continous in its domain. Therefore can we say that the function f is continous.

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

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    Last edited: Aug 29, 2007
  2. jcsd
  3. Aug 29, 2007 #2


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    What does saying that f is continuous on its domain have to do with 'continuous on an interval' 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.
  4. Aug 29, 2007 #3
    I have corrected my post. Please have a look at it again.
  5. Aug 29, 2007 #4


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

    It is not, however, "continuous on the real numbers" nor is it continuous on any interval that includes 2.
  6. Aug 29, 2007 #5

    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 continous or non-continous .
  7. Aug 29, 2007 #6
    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.
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