# Thread: Discontinuity Point Hole

1. ## Discontinuity Point Hole

Hey, I have a limit discontinuity problem here, and I think I've done it right, but I wanted to make sure that my reasoning was correct.

Question;

For $\displaystyle f(x)=\left\{\begin{array}{cc}\sqrt{x-4},&\mbox{ if } x>4\\8-2x, & \mbox{ if } x<4\end{array}\right.$, Use the mathematical definition of continuity to determine whether $\displaystyle f(x)$ is continuous at $\displaystyle x=4$.

I found the left and right hand limits to both equal 0, but the limit cannot be defined directly as x approaches 4, and the function is not defined at 4, so I think there is a point hole in the function at 4. Am I right?

Thanks for your time!

2. Originally Posted by superduper
Hey, I have a limit discontinuity problem here, and I think I've done it right, but I wanted to make sure that my reasoning was correct.

Question;

For $\displaystyle f(x)=\left\{\begin{array}{cc}\sqrt{x-4},&\mbox{ if } x>4\\8-2x, & \mbox{ if } x<4\end{array}\right.$, Use the mathematical definition of continuity to determine whether $\displaystyle f(x)$ is continuous at $\displaystyle x=4$.

I found the left and right hand limits to both equal 0, but the limit cannot be defined directly as x approaches 4, and the function is not defined at 4, so I think there is a point hole in the function at 4. Am I right?

Thanks for your time!
The left hand and right hand limits have to be equal and furthermore have to be equal to f(4). The second criterion fails because f(4) is not defined. So yes, you are correct.

3. Originally Posted by mr fantastic
The left hand and right hand limits have to be equal and furthermore have to be equal to f(4). The second criterion fails because f(4) is not defined. So yes, you are correct.
Cool thanks alot