Continuous Function Description

Question

Did I describe this functions continuity right?

The original function is $f(x)=x\sqrt{x+3}$, and the original question is in which interval is the function continuous?

My answer using set-builder notation is:

$x\in\mathbb{R}\lvert x\ge-3$

Is my answer correct?

Answer 1

The answer is $x > -3$.. Note that $x = -3$ is excluded since the function is not continuous at $x = -3$. This is so because the function is not defined when $x$ approaches $-3$ from the left (i.e. $x \rightarrow -3 - \epsilon$ for any $\epsilon > 0$.

Answer 2

A function $f$ for which $\textrm{dom}(f) \subseteq \mathbb{R}$ and $\textrm{ran}(f) \subseteq \mathbb{R}$ that can be obtained by the composition of continuous functions $f_1, \ldots, f_k$ for which $\textrm{dom}(f_i) \subseteq \mathbb{R}$ and $\textrm{ran}(f_i) \subseteq \mathbb{R}$ for all $i \in [1;k]$ will itself be continuous, but only for the $x_0 \in S$ for which $\lim_{x \rightarrow^+ x_0} f(x) = \lim_{x \rightarrow^- x_0} f(x)$.

The domain of $f(x) = x \sqrt{x+3}$ is $\{ x \in \mathbb{R} \mid x \geq -3 \}$. $f$ is therefore continuous for $\{ x \in \mathbb{R} \mid x > -3 \}$.