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.

Negative values for $x$ should be included for the function
$$h(x)=\sqrt{x^2}=|x|$$
but as noticed in this case $f(x)=\left(\sqrt {x}\right)^2$ stands for $f(x)=\sqrt x\sqrt x$ therefore negative values for $x$ are not allowed.
More precisely the given function is the composition $f=g\circ h$ of $h(x)=\sqrt x$ with $g(x)=x^2$ and by definition we first apply $h$ and  then $g$.

edit
Therefore $\lim\limits_{x \to 0} f(x)$ is defined only as right side limit that is
$$\lim\limits_{x \to 0^+} f(x)$$
Usually we simply write
$$\lim\limits_{x \to 0} \sqrt x=0$$
implicitly assuming $x\ge 0$ that is $x\to 0^+$.
