# Limits of Vector-Valued Functions

Definition: If $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function, then $\lim_{t \to a} \vec{r}(t) = \left ( \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t) \right )$ provided that the limits of the components exist. |

//Limits of vector-valued functions in $\mathbb{R}^n$ are defined similarly as the limit of each component. //

Let's look at some examples of evaluating limits of vector-valued functions. Consider the vector-valued function $\vec{r}(t) = (t^2 - 1, t + 1, e^t)$ and suppose that we wanted to compute $\lim_{t \to 2} \vec{r}(t)$. To compute this limit, all we need to do is compute the limits of the components.

(1)For another example, consider the vector-valued function $\vec{r}(t) = \left ( \frac{e^t - 1}{t}, \frac{t - 1}{t+1}, t^2 + 3 \right )$ and suppose that we wanted to compute $\lim_{t \to 0} \vec{r}(t)$. To compute this limit, we will compute all of the limits of the components again, however, this time the limits are a little trickier to compute. Fortunately, we have already learned about various rules to evaluate limits.

(2)For $\lim_{t \to 0} \frac{e^t - 1}{t}$ we will use L'Hospital's Rule, and so $\lim_{t \to 0} \frac{e^t - 1}{t} \overset{H} = \lim_{t \to 0} \frac{e^t}{1} = 1$.

For $\lim_{t \to 0} \frac{t - 1}{t+1}$, we can use direct substitution and so $\lim_{t \to 0} \frac{t-1}{t+1} = -1$.

Now $\lim_{t \to 0} t^2 + 3$ is also easy to compute by direct substitution and so $\lim_{t \to 0} t^2 + 3 = 3$.

Thus we have that $\lim_{t \to 0} \vec{r}(t) = (1, -1, 3)$.

The following theorem gives us a formal definition to say a vector-valued function $\vec{r}(t)$ has limit $\vec{b}$ at $t = a$, which is analogous to that of limits of real-valued functions.

Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function and let $\vec{b} = (b_1, b_2, b_3) \in \mathbb{R}^3$. Then $\lim_{t \to a} \vec{r}(t) = \vec{b}$ if and only if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $0 < \mid t - a \mid < \delta$ then $\| \vec{r}(t) - \vec{b} \| < \epsilon$. |

**Proof:**$\Rightarrow$ Suppose that $\lim_{t \to a} \vec{r}(t) = \vec{b}$. Then we have that:

- Now recall that two vectors are equal if and only if their components are equal, and so the equation above implies that $\lim_{t \to a} x(t) = b_1$, $\lim_{t \to a} y(t) = b_2$, and $\lim_{t \to a} z(t) = b_3$. Now notice that these three limits are limits of real-valued functions.

- Since $\lim_{t \to a} x(t) = b_1$ then $\forall \epsilon > 0$ $\exists \delta_1 > 0$ such that if $0 < \mid t - a \mid < \delta_1$ then $\mid x(t) - b_1 \mid < \frac{\epsilon}{3}$.

- Since $\lim_{t \to a} y(t) = b_2$ then $\forall \epsilon > 0$ $\exists \delta_2 > 0$ such that if $0 < \mid t - a \mid < \delta_2$ then $\mid y(t) - b_2 \mid < \frac{\epsilon}{3}$.

- Since $\lim_{t \to a} z(t) = b_3$ then $\forall \epsilon > 0$ $\exists \delta_3 > 0$ such that if $0 < \mid t - a \mid < \delta_3$ then $\mid z(t) - b_3 \mid < \frac{\epsilon}{3}$.

- Let $\delta = \mathrm{min} \{ \delta_1, \delta_2, \delta_3 \}$. Then if $0 < \mid t - a \mid < \delta$ we have that:

- $\Leftarrow$ Suppose that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $0 < \mid t - a \mid < \delta$ then $\| \vec{r}(t) - \vec{b} \| < \epsilon$. Therefore we have that $\| \vec{r}(t) - \vec{b} \| = \| (x(t) - b_1, y(t) - b_2, z(t) - b_3) \| < \epsilon$, which implies that:

- Now since all terms of the lefthand side of this equation are positive, we must have that for $0 < \mid t - a \mid < \delta$ then $(x(t) - b_1)^2 < \epsilon^2$, $(y(t) - b_2)^2 < \epsilon^2$, and $(z(t) - b_3)^2 < \epsilon^2$, and so $\mid x(t) - b_1 \mid < \epsilon$, $\mid y(t) - b_2 \mid < \epsilon$ and $\mid z(t) - b_3 \mid < \epsilon$. Therefore by the definition of real-valued function limits we have that $\lim_{t \to a} x(t) = b_1$, $\lim_{t \to a} y(t) = b_2$, and $\lim_{t \to a} z(t) = b_3$.

- Thus $\lim_{t \to a} \vec{r}(t) = \left ( \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t) \right ) = (b_1, b_2, b_3) = \vec{b}$. $\blacksquare$