If $f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2 \sin t & t & 2t \\ \sin t & t & t \end{vmatrix}$, then $\lim\limits_{t \to 0} \frac{f(t)}{t^2}$ is equal to

$0$

The correct answer is Option (1) → $0$ ##

We are given, $f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2 \sin t & t & 2t \\ \sin t & t & t \end{vmatrix}$

On expanding along $R_1$, we get

$\begin{aligned} f(t) &= \cos t [t^2 - 2t^2] - t [2t \sin t - 2t \sin t] + 1 [2t \sin t - t \sin t] \\ &= \cos t [-t^2] - 0 + [2t \sin t - t \sin t] \\ &= -t^2 \cos t + t \sin t \quad \dots(i) \end{aligned}$

On dividing Eq. (i) by $t^2$, we get

$\frac{f(t)}{t^2} = -\cos t + \frac{\sin t}{t}$

Now, $\lim\limits_{t \to 0} \frac{f(t)}{t^2} = \lim\limits_{t \to 0} (-\cos t) + \lim\limits_{t \to 0} \frac{\sin t}{t} = -1 + 1 = 0 \quad \left[ ∵\lim\limits_{x \to 0} \frac{\sin x}{x} = 1, \cos 0 = 1 \right]$