Domain of square root of x squared What is the domain of $f(x)=(\sqrt{x})^2$ ? Is it all real numbers, or are negative numbers still excluded, even after the square?
Edit: What I'm really wondering is whether $\lim\limits_{x \to 0} f(x)$ is defined. Sorry for the confusion.

You can't "determine" a domain. The domain is given. The domain of $\sqrt{x}$ is anything you want it to be. What do you want? You want only real values (as the image)? Ok, then you must decide the domain is $x \in \mathbb{R}: x \geq 0$.
I will add, since I guess this is closer to your question: Is $\left(\sqrt{-1}\right)^2 = -1$? And does the fact that (it at least appears) we can have $x = -1$ and still get a real ($\mathbb{R}$) result. So does this "expand" the domain?
The answer is absolutely not. $\sqrt{-1} = i$ (and indeed $i^2 = -1$), so if you're afraid of complex numbers then you're still not allowed to use the negative square root.
