f(x) = { x ; x<= 0

x^2 ; 0 < x < =1

2/x ; 1 < x <= 2

x-1 ; x>2

My problem is whether f(x) is continuous at x = 0 or not. Because left hand limit = f(0) when x approaches to 0 but the right hand limit f(0) when x approaches to 0. is it continuous ?

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- Apr 26th 2011, 06:02 AMwolf12Continuity
f(x) = { x ; x<= 0

x^2 ; 0 < x < =1

2/x ; 1 < x <= 2

x-1 ; x>2

My problem is whether f(x) is continuous at x = 0 or not. Because left hand limit = f(0) when x approaches to 0 but the right hand limit f(0) when x approaches to 0. is it continuous ? - Apr 26th 2011, 06:19 AMProve It
The left hand limit is 0, since the function is f(x) = x to the left of x = 0.

The right hand limit is 0, since the function is f(x) = x^2 to the right of x = 0.

Since the left hand limit and right hand limit are equal, the function is continuous at x = 0. - Apr 26th 2011, 06:22 AMTed
@ Reply #2 :

Limit from right at x=a = Limit from left x=a does not mean the function continuous at a

this means the limit at x=a exist

For the continuity, Limit at x=a must equal f(a) - Apr 26th 2011, 06:27 AMwolf12
- Apr 26th 2011, 06:34 AMProve It
- Apr 26th 2011, 07:10 AMwolf12
- Apr 26th 2011, 07:16 AMProve It
I suggest you draw the graph... A picture tells a thousand words...

- Apr 26th 2011, 07:28 AMwolf12
- Apr 26th 2011, 07:41 AMProve It
The point is that the function needs to approach the same value from both sides. If you follow the function from the left, do you get to the same point as you would if you followed the function from the right?

- Apr 26th 2011, 08:03 AMwolf12
- Apr 26th 2011, 08:07 AMProve It
Sort of. Like I said, when dealing with limits, you do not worry about what the function equals at a point. All you want to know is what the function tends to NEAR that point.

I think you can see that as you get closer and closer to x = 0 on each side, you get closer and closer to f(x) = 0... Since this happens on both sides, the limit exists, and since the function is defined at that point and is equal to the limit, the function is continuous. Does that make sense?