Answer
$i+j+k$
Work Step by Step
Find the limit for $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})$
$ \lim\limits_{t\rightarrow 0} e^{-3t}=e^{0}=1\\ \lim\limits_{t\rightarrow 0} \dfrac{t^{2}}({\sin^{2}t}=\lim\limits_{t\rightarrow 0} \dfrac{\sin t}{t})^{-2}=1\\ \lim\limits_{t\rightarrow 0} \cos 2t =\cos 0=1$
Hence, we get $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=i+j+k$
