Question Find the limit.>3. \( \lim _{t \rightarrow 0}\left(e^{-3 t} \mathbf{i}+\frac{t^{2}}{\sin ^{2} t} \mathbf{j}+\cos 2 t \mathbf{k}\right) \) 4. \( \lim _{t \rightarrow 1}\left(\frac{t^{2}-t}{t-1} \mathbf{i}+\sqrt{t+8} \mathbf{j}+\frac{\sin \pi t}{\ln t} \mathbf{k}\right) \)>5. \( \lim _{t \rightarrow \infty}\left\langle\frac{1+t^{2}}{1-t^{2}}, \tan ^{-1} t, \frac{1-e^{-2 t}}{t}\right\rangle \) 6. \( \lim _{t \rightarrow \infty}\left\langle t e^{-t}, \frac{t^{3}+t}{2 t^{3}-1}, t \sin \frac{1}{t}\right\rangle \)

\lim _{t \rightarrow 0} e^{-3 t}=e^{0}=1, \lim _{t \rightarrow 0} \frac{t^{2}}{\sin ^{2} t}=\lim _{t \rightarrow 0} \frac{1}{\frac{\sin ^{2} t}{t^{2}}}=\frac{1}{\lim _{t \rightarrow 0} \frac{\sin ^{2} t}{t^{2}}}=\frac{1}{\left(\lim _{t \rightarrow 0} \frac{\sin t}{t}\right)^{2}}=\frac{1}{1^{2}}=1and \lim _{t \rightarrow 0} \cos 2 t=\cos 0=1. Thus\begin{array}{l}\lim _{t \rightarrow 0}\left(e^{-3 t} \mathbf{i}+\frac{t^{2}}{\sin ^{2} t} \mathbf{j}+\cos 2 t \mathbf{k}\right)=\left[\lim _{t \rightarrow 0} e^{-3 t}\right] \mathbf{i}+\left[\lim _{t \rightarrow 0} \frac{t^{2}}{\sin ^{2} t}\right] \mathbf{j}+\left[\lim _{t \rightarrow 0} \cos 2 t\right] \mathbf{k}=\mathbf{i}+\mathbf{j}+\mathbf{k} . \\\lim _{t \rightarrow 1} \frac{t^{2}-t}{t-1}=\lim _{t \rightarrow 1} \frac{t(t-1)}{t-1}=\lim _{t \rightarrow 1} t=1, \lim _{t \rightarrow 1} \sqrt{t+8}=3, \lim _{t \rightarrow 1} \frac{\sin \pi t}{\ln t}=\lim _{t \rightarrow 1} \frac{\pi \cos \pi t}{1 / t}=-\pi \quad \text { [by l'Hospital's Rule]. }\end{array}Thus the given limit equals \mathbf{i}+3 \mathbf{j}-\pi \mathbf{k}.\begin{array}{l}\lim _{t \rightarrow \infty} \frac{1+t^{2}}{1-t^{2}}=\lim _{t \rightarrow \infty} \frac{\left(1 / t^{2}\right)+1}{\left(1 / t^{2}\right)-1}=\frac{0+1}{0-1}=-1, \lim _{t \rightarrow \infty} \tan ^{-1} t=\frac{\pi}{2}, \lim _{t \rightarrow \infty} \frac{1-e^{-2 t}}{t}=\lim _{t \rightarrow \infty} \frac{1}{t}-\frac{1}{t e^{2 t}}=0-0=0 . \text { Thus } \\\lim _{t \rightarrow \infty}\left\langle\frac{1+t^{2}}{1-t^{2}}, \tan ^{-1} t, \frac{1-e^{-2 t}}{t}\right\rangle=\left\langle-1, \frac{\pi}{2}, 0\right\rangle . \\\lim _{t \rightarrow \infty} t e^{-t}=\lim _{t \rightarrow \infty} \frac{t}{e^{t}}=\lim _{t \rightarrow \infty} \frac{1}{e^{t}}=0 \quad \text { [by l'Hospital's Rule], } \lim _{t \rightarrow \infty} \frac{t^{3}+t}{2 t^{3}-1}=\lim _{t \rightarrow \infty} \frac{1+\left(1 / t^{2}\right)}{2-\left(1 / t^{3}\right)}=\frac{1+0}{2-0}=\frac{1}{2}, \\\text { and } \lim _{t \rightarrow \infty} t \sin \frac{1}{t}=\lim _{t \rightarrow \infty} \frac{\sin (1 / t)}{1 / t}=\lim _{t \rightarrow \infty} \frac{\cos (1 / t)\left(-1 / t^{2}\right)}{-1 / t^{2}}=\lim _{t \rightarrow \infty} \cos \frac{1}{t}=\cos 0=1 \quad \text { [again by l'Hospital's Rule]. } \\\text { Thus } \lim _{t \rightarrow \infty}\left\langle t e^{-t}, \frac{t^{3}+t}{2 t^{3}-1}, t \sin \frac{1}{t}\right\rangle=\left\langle 0, \frac{1}{2}, 1\right\rangle .\end{array} ...