Answer
$\left\langle {2,\frac{1}{2},\sin 2} \right\rangle $
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 1} \left\langle {\frac{3}{{{t^2}}},\frac{{\ln t}}{{{t^2} - 1}},\sin 2t} \right\rangle \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{t \to 1} \left\langle {\frac{3}{{{t^2}}},\frac{{\ln t}}{{{t^2} - 1}},\sin 2t} \right\rangle = \left\langle {\mathop {\lim }\limits_{t \to 1} \frac{3}{{{t^2}}},\mathop {\lim }\limits_{t \to 1} \frac{{\ln t}}{{{t^2} - 1}},\mathop {\lim }\limits_{t \to 1} \sin 2t} \right\rangle \cr
& {\text{Where }}\mathop {\lim }\limits_{t \to 1} \frac{{\ln t}}{{{t^2} - 1}} = \frac{0}{0} \cr
& {\text{By the L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{t \to 1} \frac{{\ln t}}{{{t^2} - 1}} = \mathop {\lim }\limits_{t \to 1} \frac{{\frac{d}{{dt}}\left[ {\ln t} \right]}}{{\frac{d}{{dt}}\left[ {{t^2} - 1} \right]}} = \mathop {\lim }\limits_{t \to 1} \frac{{1/t}}{{2t}} = \mathop {\lim }\limits_{t \to 1} \frac{1}{{2{t^2}}} = \frac{1}{2} \cr
& {\text{Therefore,}} \cr
& \left\langle {\mathop {\lim }\limits_{t \to 1} \frac{3}{{{t^2}}},\mathop {\lim }\limits_{t \to 1} \frac{{\ln t}}{{{t^2} - 1}},\mathop {\lim }\limits_{t \to 1} \sin 2t} \right\rangle = \left\langle {\frac{3}{{{{\left( 1 \right)}^2}}},\frac{1}{2},\sin 2\left( 1 \right)} \right\rangle \cr
& {\text{ }} = \left\langle {3,\frac{1}{2},\sin 2} \right\rangle \cr} $$
