## Calculus: Early Transcendentals 8th Edition

$i+j+k$
Given: $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=(1)\mathrm{i}+(1)\mathrm{j}+(1) \mathrm{k}$ $\lim\limits_{t\rightarrow 0} e^{-3t}=e^{0}=1$ and $\lim\limits_{t\rightarrow 0} \dfrac{t^{2}}{\sin^{2}t}=\lim\limits_{t\rightarrow 0} (\dfrac{\sin t}{t})^{-2}=[\lim\limits_{t\rightarrow 0} \dfrac{\sin t}{t}]^{-2}=1$ and $\lim\limits_{t\rightarrow 0} \cos 2t =\cos (0)=1$ $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=(1) \times \mathrm{i}+(1) \times \mathrm{j}+(1) \times \mathrm{k}=i+j+k$ Hence, we have $i+j+k$