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Continuity of sqrt(x) at x = 0

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    The question is to find 2 functions (f(x) and g(x) let's say) such that they're both NOT continuous at point a but at the same time, f(x)+g(x) and f(x)g(x) are continuous.

    2. Relevant equations

    3. The attempt at a solution
    I was thinking of letting f(x) = x + [tex]\sqrt{x}[/tex] and g(x) = x - [tex]\sqrt{x}[/tex], claiming that f(x) and g(x) are not continuous at a = 0. This yields f(x) + g(x) = 2x and f(x)g(x) = [tex]x^{2} - x[/tex]. However, that is the problem at hand. Is [tex]\sqrt{x}[/tex] continuous at x = 0? Using the definition of continuity, the limit does NOT exist as you can only find the limit on one-side (not considering the complex plane). However, according to my textbook (Stewart), it says that all root functions are continuous for every number in its domain. If the latter is the case, what two functions would satisfy the above? Thank you so much for your help guys!
  2. jcsd
  3. Oct 18, 2007 #2


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    Roots are continuous. Sqrt(x) is defined for x > 0, so the left limit is not applicable at x = 0.
  4. Oct 18, 2007 #3
    here's a hint : mod functions.
  5. Oct 18, 2007 #4


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    Think of the functions graphically and what discontinuous functions look like. Draw a bunch of different kinds and think of how you maybe be able to add them together and piece them together to make them continuous after adding them.
  6. Oct 18, 2007 #5
    Well my original tactic was to let
    f(x) = x + (some discontinuous function)
    g(x) = x - (some discontinuous function)
    so that f(x) + g(x) = 2x and f(x)g(x) = [tex]x^{2}[/tex] - (some discontinuous function)[tex]^{2}[/tex] hoping that the latter would become continuous once squared (which is why I wondered if [tex]\sqrt{x}[/tex] was discontinuous at 0 or not). But since that isn't the case, I guess I've got to find some other way.
    Last edited: Oct 18, 2007
  7. Oct 19, 2007 #6
    What kind of discontinuities can you add so that it'd produce a continuous function? The only way I see is to get rid of them by cancelling them (hence my previous post). I thought about floor and ceiling functions as someone suggested but what can you add to them to make it continuous o_O.
  8. Oct 19, 2007 #7


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    How about step functions?
  9. Oct 20, 2007 #8
    Hey, here is an example :
    f(x) = x + |x|
    g(x) = x - |x|
  10. Oct 20, 2007 #9
    Aren't those continuous in the first place? f(x) is continuous for all x in its domain and is right-continuous at x = 0 and g(x) is also continuous with it being left-continuous at x = 0. Otherwise, wouldn't my example with f(x) = x - sqrt(x) and g(x) = x + sqrt(x) have worked?
  11. Oct 20, 2007 #10


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    The sqrt thing sort of works, but for the wrong reason. Think of f(x)=1 if x>=0 and f(x)=0 if x<0. Let g(x)=1-f(x).
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