Thank you for reporting, we will resolve it shortly
Q.
If $f(t)=\begin{vmatrix}\cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t\end{vmatrix}$, then $\displaystyle\lim _{t \rightarrow 0} \frac{f(t)}{t^2}$ is equal to
Determinants
Solution:
Consider, $f(t)=\begin{vmatrix}\cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t\end{vmatrix}$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$,
$\Rightarrow f(t)=\begin{vmatrix} \cos t & t & 1 \\2 \sin t-\cos t & 0 & 2 t-1 \\\sin t-\cos t & 0 & t-1\end{vmatrix}$
By expanding along $C _2$, we get
$f(t)=-t \begin{vmatrix} 2 \sin t-\cos t & 2 t-1 \\ \sin t-\cos t & t-1\end{vmatrix}$
$=(-t)[(t-1)(2 \sin t-\cos t)-(2 t-1)(\sin t-\cos t)]$
$=(-t)[2 \sin t \cdot t-\cos t \cdot t-2 \sin t+\cos t$
$-2 t \cdot \sin t+2 t \cdot \cos t+\sin t-\cos t]$
$\Rightarrow f(t)=(-t)[t \cdot \cos t-\sin t]$
$\Rightarrow f(t)=t \sin t-t^2 \cos t$
$\therefore \displaystyle\lim _{t \rightarrow 0} \frac{f(t)}{t^2}=\displaystyle\lim _{t \rightarrow 0} \frac{t \sin t-t^2 \cos t}{t^2}$
$=\displaystyle\lim _{t \rightarrow 0} \frac{\sin t}{t}-\displaystyle\lim _{t \rightarrow 0} \cos t$
$=1-1=0$