Answer
$\mathop {\lim }\limits_{t \to 0} \frac{{{\bf{r}}\left( t \right)}}{t} = \left( {1,0, - 2} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\sin t,1 - \cos t, - 2t} \right)$.
By Theorem 1, vector-valued limits are computed component-wise, thus
$\mathop {\lim }\limits_{t \to 0} \frac{{{\bf{r}}\left( t \right)}}{t} = \left( {\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t},\mathop {\lim }\limits_{t \to 0} \frac{{1 - \cos t}}{t},\mathop {\lim }\limits_{t \to 0} \left( { - 2} \right)} \right)$
1. Evaluate $\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}$.
Using L'Hôpital's rule we get
$\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = \mathop {\lim }\limits_{t \to 0} \frac{{\cos t}}{1} = 1$
2. Evaluate $\mathop {\lim }\limits_{t \to 0} \frac{{1 - \cos t}}{t}$
Using L'Hôpital's rule we get
$\mathop {\lim }\limits_{t \to 0} \frac{{1 - \cos t}}{t} = \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{1} = 0$
3. Evaluate $\mathop {\lim }\limits_{t \to 0} \left( { - 2} \right)$
$\mathop {\lim }\limits_{t \to 0} \left( { - 2} \right) = - 2$
So, $\mathop {\lim }\limits_{t \to 0} \frac{{{\bf{r}}\left( t \right)}}{t} = \left( {1,0, - 2} \right)$.
