# Discontinuity...again

• May 31st 2009, 01:49 AM
Showcase_22
Discontinuity...again
Is $\displaystyle f(x)=[ \sqrt{x} ]$ where $\displaystyle f:[0, \infty) \rightarrow \mathbb{R}$ is continuous at 0?

I put false since I can see that the function is not left continuous so it cannot be continuous.

is this right?
• May 31st 2009, 02:12 AM
HallsofIvy
Quote:

Originally Posted by Showcase_22
Is $\displaystyle f(x)=[ \sqrt{x} ]$ where $\displaystyle f:[0, \infty) \rightarrow \mathbb{R}$ is continuous at 0?

I put false since I can see that the function is not left continuous so it cannot be continuous.

is this right?

I assume that you meant simply $\displaystyle f(x)= \sqrt{x}$. We only require that $\displaystyle \lim_{x\rightarrow a} f(x)= f(a)$ for x in the domain of f in order that f be continuous at a. The domain of $\displaystyle \sqrt{x}$ is all non-negative x. So, no, the fact that $\displaystyle \sqrt{x}$ is not defined for negative x does not prevent it being continuous at x= 0.
• May 31st 2009, 02:28 AM
Showcase_22
I did actually mean $\displaystyle f(x)=[\sqrt{x}]$

By your idea the function is continuous at 0 because it's sequentially continuous?