Find the limit of the vector as $~t~$ approaches $~0~$? Here's the vector: $$e^{-6t}~\vec i + \frac{t^2}{\sin^2t}~\vec j + \sin(6t)~\vec k$$
Don't I just take the limit as $t$ approaches $0$ from each individual component? 
As a result, I got $~\vec i + \vec j~$, which was apparently incorrect.
 A: You are correct, we just take the limit of each component for vector valued functions.
The $e^{-6t}$ term clearly goes to 1 by substitution.  
For the second term we can use L'Hôpital's rule (there is probably a more elegant way). 
$$\lim _{t\to0}\frac{t^2}{\sin^2(t)}=\lim_{t\to0}\frac{t}{\sin t\cos t}$$
$$=\lim_{t\to 0}\frac{t}{\sin t}\cdot\lim_{t\to0}\frac{1}{\cos t}$$
$$=1$$
The third term is also solved by simple substitution and goes to 0. 
So I think your result is correct. 
$$\lim_{t\to0}\left\langle e^{-6t}, \frac{t^2}{\sin^2t}, \sin(6t) \right\rangle=\langle 1, 1, 0 \rangle=\hat\imath+\hat\jmath$$
A: $$\lim_{t\to 0} \left\{e^{-6t}~\vec i + \frac{t^2}{\sin^2t}~\vec j + \sin(6t)~\vec k\right\}$$
$$=\left\{\lim_{t\to 0}  ~e^{-6t}~\right\}~\vec i + \left\{\lim_{t\to 0}  ~\frac{t^2}{\sin^2t}~\right\}~\vec j + \left\{\lim_{t\to 0}  ~ \sin(6t)~\right\}~\vec k $$
$$=e^0~\vec i ~+~\vec j~+~\sin(0)~\vec k$$
$$=\vec i ~+~\vec j$$



*

*$$~\lim_{t\to 0}  ~\frac{t^2}{\sin^2t}=~\lim_{t\to 0}  \left(~\frac{t}{\sin t}\right)^2=\left(~\lim_{t\to 0}~\frac{t}{\sin t}\right)^2=\frac{1}{\lim_{t\to 0}~\frac{\sin t}{t}}=1~$$
