[Xatima]

Is the following function continuous at x=1?if yes,HOW?

f(x)= x-|x|

[silent ninja]

Post in Maths help.

Draw a sketch.

[Diomedes]

Originally Posted by Xatima

Is the following function continuous at x=1?if yes,HOW?

f(x)= x-|x|

Are you required to use the definition or can you use limits?

[Xatima]

Originally Posted by Diomedes

Are you required to use the definition or can you use limits?

either way.

[Slumpy]

Originally Posted by Xatima

either way.

Consider f(x) on (0,infinity) then.
What do you know about f?

[Xatima]

but i am supposed to check continuity at 1.

[Slumpy]

Yes, but there's a fact about f(x) on that entire interval that will really help.

[Xatima]

at 0 its continuous
but at infinity it not.
mmmmm.....right?

[Slumpy]

I'm not talking about continuity yet, I'm just asking what you can say about f(x) for positive x?
(And, it doesn't really mean anything to say continuous at infinity, as continuity id defined pointwise, and infinity isn't a point.)

[Xatima]

F(x) for +ve x is 0

[Slumpy]

Ok, so what's limf(x) as x---> any positive number?
Is this equal to f(x)? If so, it's continuous, if not, it's not.

[Xatima]

ohhhhhhhh ok.....from 0 to infinity it's value remains same which is zero.....
sorry for eating ur brain Ah,i got ur point now.

[Xatima]

Originally Posted by Slumpy

Ok, so what's limf(x) as x---> any positive number?
Is this equal to f(x)? If so, it's continuous, if not, it's not.

yes...both are equal even if i consider limf(x) as x--->1.
BUt what is i consider it from the -ve side?does it still remain same keeping i view fact that i am considering abs function?

[Xatima]

i mean...
if i approach from -ve side,
i say it like that,since i am approaching from -ve side,i would place a -ve sign outside abs function like this
Lim x--->1-
F(x)=x-{-(x)}.I have placed - sign outside abs function while approaching from -ve side bcz i wanted to keep it's value +ve,which is definition of abs function.

[Slumpy]

The limit x--->1- is still 0.
It's not too tricky to prove rigorously, but vaguely, if you've got a sequence tending to 1 from below, there's an infinite number of points between 0 and 1, so after a certain N, all xn are 0, so the sequence converges to 0, which is f(1).