Answer
$\mathrm{i}+\mathrm{j}+\mathrm{k}$
Work Step by Step
Component by component,
$\displaystyle \lim_{t\rightarrow 0}e^{-3t}=e^{0}=1,$
$\displaystyle \lim_{t\rightarrow 0}\frac{t^{2}}{\sin^{2}t}=\lim_{t\rightarrow 0}(\frac{\sin t}{t})^{-2}=(\lim_{t\rightarrow 0}\frac{\sin t}{t})^{-2}=1^{-2}=1$,
$\displaystyle \lim_{t\rightarrow 0}\cos 2t =\cos 0=1$.
$\displaystyle \lim_{t\rightarrow 0} (e^{-3t}$i$+\displaystyle \frac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=(1)\mathrm{i}+(1)\mathrm{j}+(1) \mathrm{k}\\=\mathrm{i}+\mathrm{j}+\mathrm{k}$.
