Continuity of sqrt(x) at x = 0

  • Thread starter bjgawp
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In summary: Then g(x)=-1 for all x>=0 but is 1 for all x<0. So g(x) is discontinuous at x>=0 but continuous at x<0.
  • #1
bjgawp
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Homework Statement


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.



Homework Equations





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!
 
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  • #2
Roots are continuous. Sqrt(x) is defined for x > 0, so the left limit is not applicable at x = 0.
 
  • #3
here's a hint : mod functions.
 
  • #4
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.
 
  • #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.
 
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  • #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.
 
  • #7
How about step functions?
 
  • #8
Hey, here is an example :
f(x) = x + |x|
g(x) = x - |x|
 
  • #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?
 
  • #10
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).
 

Related to Continuity of sqrt(x) at x = 0

What is the definition of continuity?

Continuity is a property in mathematics that describes the behavior of a function at a specific point or interval. A function is considered continuous if it has no breaks or gaps in its graph and if its limit exists at that point or interval.

How is continuity of a function determined at a specific point?

To determine the continuity of a function at a specific point, we must check three conditions: 1) the function is defined at that point, 2) the limit of the function exists at that point, and 3) the limit is equal to the value of the function at that point. If all three conditions are met, the function is continuous at that point.

Is the function sqrt(x) continuous at x = 0?

Yes, the function sqrt(x) is continuous at x = 0. This is because the function is defined at x = 0, the limit of the function exists at x = 0 and is equal to 0, and the value of the function at x = 0 is also 0. Therefore, all three conditions for continuity are met at x = 0.

Can a function be continuous at a point but not on an interval?

Yes, a function can be continuous at a point but not on an interval. This means that the function is continuous at that specific point, but it may have breaks or gaps in its graph on the interval surrounding that point. This can occur if the limit at that point is not equal to the value of the function at that point.

How is the continuity of sqrt(x) at x = 0 related to the concept of differentiability?

The continuity of a function is a necessary condition for it to be differentiable at that point. In the case of sqrt(x) at x = 0, since the function is continuous at x = 0, it is also differentiable at that point. This means that the graph of sqrt(x) has a smooth and continuous tangent line at x = 0.

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