Answer
$\mathop {\lim }\limits_{h \to 0} \frac{{{\bf{r}}\left( {t + h} \right) - {\bf{r}}\left( t \right)}}{h} = \left( { - {t^{ - 2}},\cos t,0} \right)$
Work Step by Step
By definition, the derivative of ${\bf{r}}\left( t \right)$ as the limit of the difference quotient is
${\rm{r}}'\left( t \right) = \frac{d}{{dt}}{\bf{r}}\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{\bf{r}}\left( {t + h} \right) - {\bf{r}}\left( t \right)}}{h}$
By Theorem 2, vector-valued derivatives are computed component-wise, therefore
${\rm{r}}'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{\bf{r}}\left( {t + h} \right) - {\bf{r}}\left( t \right)}}{h} = \left( {x'\left( t \right),y'\left( t \right),z'\left( t \right)} \right)$
We have ${\bf{r}}\left( t \right) = \left( {{t^{ - 1}},\sin t,4} \right)$. So, the coordinate functions are
$x\left( t \right) = {t^{ - 1}}$, ${\ \ }$ $y\left( t \right) = \sin t$, ${\ \ }$ $z\left( t \right) = 4$
Thus,
$\mathop {\lim }\limits_{h \to 0} \frac{{{\bf{r}}\left( {t + h} \right) - {\bf{r}}\left( t \right)}}{h} = \left( {x'\left( t \right),y'\left( t \right),z'\left( t \right)} \right)$
$\mathop {\lim }\limits_{h \to 0} \frac{{{\bf{r}}\left( {t + h} \right) - {\bf{r}}\left( t \right)}}{h} = \left( { - {t^{ - 2}},\cos t,0} \right)$
