If $f(t) = \left[ {\begin{array}{cccccccccccccccccccc}{\cos t}&t&1\\{2\sin t}&t&{2t}\\{\sin t}&t&t\end{array}} \right]$, then $\mathop {\lim }\limits_{t \to 0} \frac{{f(t)}}{{{t^2}}}$ is equal to

We have
$f(t) = \left| {\begin{array}{cccccccccccccccccccc}{\cos t}&t&1\\{2\sin t}&t&{2t}\\{\sin t}&t&t\end{array}} \right|$

Expanding along ${C_1}$,

$= \cos t\left( {{t^2} - 2{t^2}} \right) - 2\sin t\left( {{t^2} - t} \right) + \sin t\left( {2{t^2} - t} \right)$

$= - {t^2}\cos t - \left( {{t^2} - t} \right)2\sin t + \left( {2{t^2} - t} \right)\sin t$

$= - {t^2}\cos t - {t^2} \cdot 2\sin t + t \cdot 2\sin t + 2{t^2}\sin t$

$= - {t^2}\cos t + 2t\sin t$

$\therefore \mathop {\lim }\limits_{t \to 0} \frac{{f(t)}}{{{t^2}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\left( { - {t^2}\cos t} \right)}}{{{t^2}}} + \mathop {\lim }\limits_{t \to 0} \frac{{2t\sin t}}{{{t^2}}}$

$= - \mathop {\lim }\limits_{t \to 0} \cos t + 2 \cdot \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}$

$= - 1 + 1$
$= 0$