Continuity of Vector-Valued Functions

Like with continuity of single variable real-valued functions, the continuity of vector-valued functions is as follows:

Definition Let $\vec{r}(t)$ be a vector-valued function. We say that $\vec{r}(t) = (x(t), y(t), z(t))$ is Continuous at $a$ if $\lim_{t \to a} \vec{r}(t) = \vec{r}(a)$, that is $\lim_{t \to a} x(t) = x(a)$, $\lim_{t \to a} y(t) = y(a)$ and $\lim_{t \to a} z(t) = z(a)$. We say that $\vec{r}(t)$ is Continuous on The Interval $I$ if $\vec{r}(t)$ is continuous for all $t \in I$.

We should note that from the definition of $\vec{r}(t)$ being continuous at $a$, we must have that the component functions $x(t)$, $y(t)$, and $z(t)$ all be continuous at $a$.

For example, suppose we wanted to show that $\vec{r}(t) = (t, -2t^2, \cos t)$ is continuous at $\pi$. We notice that $x(t) = t$, $y(t) = -2t^2$, and $z(t) = \cos t$ are all continuous real-valued functions, and so they're also all continuous at $\pi$. Therefore, we have that:

(1)

\begin{align} \quad \quad \lim_{t \to \pi} \vec{r}(t) = \left ( \lim_{t \to \pi} t, \lim_{t \to \pi} -2t^2, \lim_{t \to \pi} \cos t \right ) = (\pi, -2\pi^2, -1) = \vec{r}(\pi) \end{align}