Limits of Vector-Valued Functions Examples 1

Limits of Vector-Valued Functions Examples 1

Let $\vec{r}(t)$ be a vector-valued function. Recall that we can compute limit at $t \to a$ of vector-valued functions by taking the limits as $t \to a$ of each of the components respectively. We will now look at some examples of calculating some vector-valued function limits.

Example 1

Evaluate $\lim_{t \to 1} \left ( \frac{t^2 - t}{t - 1} - t , \ln 2t, 2t - t + 1 \right )$.

We will first calculate the limits of each component respectively. First, $\lim{t \to 1} \frac{t^2 - t}{t - 1} - t= \lim{t \to 1} \frac{t(t-1)}{t-1} - t = \lim{t \to 1} t - t = 0$. Secondly, $\lim{t \to 1} \ln 2t = \ln 2$. Lastly, $\lim_{t \to 1} 2t - t + 1 = 2$. Therefore we have that:

(1)
\begin{align} \quad \lim_{t \to 1} \left ( \frac{t^2 - t}{t - 1} - t , \ln 2t, 2t - t + 1 \right ) = (0, \ln 2, 2) \end{align}

Example 2

Evaluate $\lim_{t \to 0} \left ( e^{4t} \vec{i} + \frac{t^2}{\sin ^2 t} \vec{j} + \sin 3t \vec{k} \right )$.

First, $lim_{t \to 0} e^{4t} = 1$. Secondly, $\lim_{t \to 0} \frac{t^2}{\sin ^2 t} \overset{LH} = \lim_{t \to 0} \frac{2t}{2 \sin t \cos t} = \lim_{t \to 0} \frac{t}{\sin t \cos t} \overset{H} = \lim_{t \to 0} \frac{1}{\cos ^2 t - \sin ^2 t} = 1$. Lastly, $\lim_{t \to 0} \sin 3t = 0$. Therefore we have that:

(2)
\begin{align} \quad \lim_{t \to 0} \left ( e^{4t} \vec{i} + \frac{t^2}{\sin ^2 t} \vec{j} + \sin 3t \vec{k} \right ) = \vec{i} + \vec{j} + 0\vec{k} \end{align}

Example 3

Evaluate $\lim_{t \to -1} \left ( t^2 - t, \cos e^t, \frac{t^2 +2t + 1}{t + 1}, \sqrt{t^2 - t}, \cos t \right )$.

First, $\lim_{t \to -1} t^2 - t = 2$. Secondly, $\lim_{t \to -1} \cos e^t = \cos \frac{1}{e}$. Thirdly, $\lim_{t \to -1} \frac{t^2 + 2t + 1}{t + 1} = \lim_{t \to -1} \frac{(t + 1)(t + 1)}{t + 1} = \lim_{t \to -1} t + 1 = 0$. Fourthly, $\lim_{t \to -1} \sqrt{t^2 - t} = \sqrt{2}$. Lastly, $\lim_{t \to -1} \cos t = \cos (-1)$. Therefore we have that:

(3)
\begin{align} \quad \lim_{t \to -1} \left ( t^2 - t, \cos e^t, \frac{t^2 +2t + 1}{t + 1}, \sqrt{t^2 - t}, \cos t \right ) = (2, \cos \frac{1}{e}, 0, \sqrt{2}, \cos (-1) ) \end{align}
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