Continuity of sqrt(x) at x = 0

[bjgawp]

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 + \sqrt{x} and g(x) = x - \sqrt{x}, 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) = x^{2} - x. However, that is the problem at hand. Is \sqrt{x} 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!

 

[EnumaElish]

Roots are continuous. Sqrt(x) is defined for x > 0, so the left limit is not applicable at x = 0.

 

[Sourabh N]

here's a hint : mod functions.

 

[JasonRox]

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.

 

[bjgawp]

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) = x^{2} - (some discontinuous function)^{2} hoping that the latter would become continuous once squared (which is why I wondered if \sqrt{x} was discontinuous at 0 or not). But since that isn't the case, I guess I've got to find some other way.

 

[bjgawp]

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 .

 

[Dick]

How about step functions?

 

[Sourabh N]

Hey, here is an example :
f(x) = x + |x|
g(x) = x - |x|

 

[bjgawp]

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?

 

[Dick]

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).