Question

14.1 vector - valued functions not started created about 2 hours ago · last updated about 2 hours ago items turn in homework due sep 15, 2025 11:59 pm. 2. submit answer get help practice similar let $mathbf{r}(t)=(sqrt{t + 3})mathbf{i}+left(\frac{t^{2}-9}{t + 3}
ight)mathbf{j}+sin(-2pi t)mathbf{k}$. then $lim_{t
ightarrow1}mathbf{r}(t)=mathbf{i}+mathbf{j}+mathbf{k}$.

Explanation:

Step1: Find limit of i - component

For the i - component $f(t)=\sqrt{t + 3}$, we substitute $t = 1$ into it. $\lim_{t
ightarrow1}\sqrt{t + 3}=\sqrt{1+3}=\sqrt{4}=2$.

Step2: Find limit of j - component

For the j - component $g(t)=\frac{t^{2}-9}{t + 3}=\frac{(t + 3)(t - 3)}{t+3}=t - 3$ (for $t
eq - 3$). Then $\lim_{t
ightarrow1}(t - 3)=1-3=-2$.

Step3: Find limit of k - component

For the k - component $h(t)=\sin(-2\pi t)$, we substitute $t = 1$ into it. $\lim_{t
ightarrow1}\sin(-2\pi t)=\sin(-2\pi)=0$.

Answer:

$2\mathbf{i}-2\mathbf{j}+0\mathbf{k}$